You got a sequence of 1x1 matrices!
Assign that sequence to L.
Then
V:=Vector([L]);
will be a column vector with the elements from the 1x1 matrices.
Now export with ExportVector or with ExcelTools:-Export.

I don't quite see in your code any attempt to implement the piecewise defined A at the top. Presumably (t,365) is meant as t-floor(t/365)*365.

Knowing nothing about stochastic differential equations the following should be taken as an experiment.
The example used is only somewhat similar to yours.

Added: A somewhat simpler looking way at the bottom.

restart;
#The ode considered is
ode:=diff(U(t),t)=-(Ap(t)+r(t)+B*U(t))*U(t);
#where Ap is a piecewise defined function, B is a constant, and r is a random variable here defined as:

r:=RandomTools:-Generate(distribution(Normal(0, .5)), makeproc=true);
#We shall not use ode at all, but use procedural input to dsolve/numeric:

p:=proc(N,t,Y,YP) local Ap,B;
   B:=.1;
   Ap:=piecewise(t<=1,.1+ 0.4*t,0.5+ 0.004*t); 
   YP[1]:=-(Ap+r(t)+B*Y[1])*Y[1]
end proc;
#It appears to me that the only reasonable method is a forward method with no error corrections.
#Thus I use the simple Euler method classical[foreuler].
#Choose some stepsize consistent with the application we have in mind.
#Here chosen arbitrarily to 0.01.
res:=dsolve(numeric,procedure=p,number=1,initial=Array([1]),start=0,procvars=[U(t)],method=classical[foreuler],stepsize=0.01);
plots:-odeplot(res,[t,U(t)],0..2);

#########################################
##Another way and I guess simpler looking.
restart;
ode:=diff(U(t),t)=-(Ap(t)+r(t)+B*U(t))*U(t);
R:=RandomTools:-Generate(distribution(Normal(0,.5)), makeproc=true);
r:=proc(t)
   if not type(t,numeric) then return 'procname(_passed)' end if;
   R()
end proc;
Ap:=t->piecewise(t<=1,.1+ 0.4*t,0.5+ 0.004*t);
B:=0.1:
res:=dsolve({ode,U(0)=1},numeric,known=r(t),method=classical[foreuler],stepsize=0.01);
plots:-odeplot(res,[t,U(t)],0..2);

Taking jj=2 and keeping the rather large g:

soln:=dsolve({initcondNORMAL, op(subs(k = 1, n = 2, g = 10, [eq1,eq2,eq3]))}, numeric,output=listprocedure);
Z,Z1,X,X1:=op(subs(soln,[z(t),diff(z(t),t),x(t),diff(x(t),t)]));
plots:-odeplot(soln,[t,z(t)],0..0.15);
fsolve(Z1); #The time for maximal z
Z(%); #The value of zmax
T:=fsolve(Z,.15); #The time for z = 0.
arctan(Z1(T)/X1(T)); #The angle in radians
%*180/evalf(Pi); #The angle in degrees
####################
Try also

solE:=dsolve({initcondNORMAL, op(subs(k = 1, n = 2, g = 10, [eq1,eq2,eq3]))}, numeric,events=[[diff(z(t),t),halt],[z(t),halt]]);
solE(.1);
solE(eventclear);
solE(.2);

If you can live with a parenthesis aound the exponent then beta=10^``(-6) is a solution.

A rather weird way:

ten:=convert(10,symbol);
ten^(-6); #Doesn't do it, but:
series(ten^(-6),ten);

Although your example is simple, it is a polynomial, so fsolve returns 2 solutions (if they are real).
So I have taken the first one found (the [1] attached to fsolve):
p:=(a,b)->fsolve( x^2 + a*x + b =0, x)[1];
#Use argument free syntax:
plot3d( p , 0.1..5, 0.5..3,axes=boxed,labels=[a,b,p]);

When you don't have a polynomial fsolve only returns one solution, so omit the [1].
Example:

p:=(a,b)->fsolve( x + a*ln(x) + b =0, x);
plot3d( p , 0.1..5, 0.5..3,axes=boxed,labels=[a,b,p]);

You may want to take a look at the events option:

?dsolve,events

restart;
sys:=diff(x(t),t)= x(t)+y(t) , diff(y(t),t)= x(t)-y(t);  
res:=dsolve({sys,x(0)=1/2,y(0)=0},numeric,events=[[x(t)-1,halt],[y(t)-2,halt]]);
plots:-odeplot(res,[x(t),y(t)],0..5);
plots:-odeplot(res,[t,x(t)],0..5);

Here is a simple example, which probably just shows that I don't understand your problem:

restart;
with(LinearAlgebra):
M:=<seq(m[i](t),i=1..3)>;
H:=<M[2],M[1],M[3]>; #Simple example
eqs:=diff~(M,t)+ M &x H-a*M &x diff~(M,t)/DotProduct(M,M,conjugate=false)=~ <0,0,0>;
sys:={seq(eqs[i],i=1..3)};
res:=dsolve(eval(sys,a=1) union {m[1](0)=1,m[2](0)=0,m[3](0)=0}, numeric);
plots:-odeplot(res,[seq([t,m[i](t)],i=1..3)],0..10);

Here is a simple example:

v:=Vector(datatype=float);
for i from 1 to 10 do v(i):=subs(Optimization:-NLPSolve(sin(x)*x^i, x=1..5,method=branchandbound,maximize)[2], x) end do:
v;

Then use ExportVector.

In your infinity version there appear factors beta(beta), they should just be beta. See comment at the end, though.
Specifically, you have

1/c1 = 6.274557048*10^8*(Int((32.*sin(0.5000000000e-1*beta)/beta+(960.*(cos(0.5000000000e-1*beta)-40.*sin(0.5000000000e-1*beta)/beta+1600.*sin(0.2500000000e-1*beta)^2/beta^2))/beta^2)^2/((75*(1.50*tanh(beta)*beta(beta)+1))/(75+0.2e-1*tanh(beta)*beta(beta))+2.32*coth(beta)*beta(beta)), beta = 0 .. infinity));
#You see the appearance of beta(beta).
#Let A be the integral (using Int):

A:=Int((32.*sin(0.5000000000e-1*beta)/beta+(960.*(cos(0.5000000000e-1*beta)-40.*sin(0.5000000000e-1*beta)/beta+1600.*sin(0.2500000000e-1*beta)^2/beta^2))/beta^2)^2/((75*(1.50*tanh(beta)*beta(beta)+1))/(75+0.2e-1*tanh(beta)*beta(beta))+2.32*coth(beta)*beta(beta)), beta = 0 .. infinity);
#Now do
subs(beta(beta)=beta,A);
B:=normal(%);
#This integral can be evaluated, but it is not all that easy. The problem is not at beta = 0.
f:=IntegrationTools:-GetIntegrand(B);
asympt(f,beta, 9);
series(f,beta=0, 9);
plot(f,beta=0..100);
evalf(Int(f,beta=0..1000));
evalf(Int(f,beta=0..5000));

More work could be done on this, of course. You could try something like:

evalf[15](Int(f,beta=0..10000,method = _d01akc));
#and
evalf[15](Int(f,beta=0..200000,method = _d01akc,epsilon=1e-10));
#for large values of beta f is bounded from above by
g:=eval(f,{sin=1,cos=1,tanh=1,coth=1,-38400.=38400});
#so you can get an upper bound for the error by computing:
evalf[15](Int(g,beta=200000..infinity));

Comment on beta(beta):

Try the following
restart;
evalf(Int(beta(beta),beta=0..1));
evalf(Int(beta,beta=0..1));
int(beta(beta),beta=0..1);
#The reason the numerical integration works is that in Maple the result of
5(5);
#is 5, as is
5(x);

restart;
g:=y*(1-1/(x^2+y^2)):
# g = 0.1
plots:-contourplot(g,x=-1..1,y=0..2,contours=[0.1]); p:=%:
dt:=plottools:-getdata(p):
L:=ListTools:-Flatten(select~(type,[dt],Matrix)):
M:=<op(L)>;
#plot(M);
#M[1..10,..];
#Lots of repeats, so:
Mu:=Matrix(ListTools:-MakeUnique(convert(M,listlist)));
#plot(Mu);

Then do the same for g = 0 and save in Mu0, say. Notice now that plot(Mu0); connects points in a confusing manner. But you can do plot(Mu0,style=points) to avoid seeing these connecting lines.
To export Mu and Mu0 to Excel see ?ExcelTools

phi[i], i=1..N are not defined as functions (procedures) so should not be used as such. 

restart;
N:=4:
for i from 1 to N do
 phi[i]:=piecewise(t>=x[i-1] and x[i]>t, (t-x[i])/(h), x[i]<t and t<=x[i+1], (x[i+1]-t)/(h), 0);
end do;
V:=Int(diff(phi[1], t)*diff(phi[1], t), t=x[0]..x[2]);
value(V) assuming x[0]<x[1],x[1]<x[2];

Since log(y) - > -infinity as y - > 0+, there wouldn't be room on your screen for your plot. So a compromise is made.

Try for comparison

plots:-loglogplot(f, x = 0 .. 5*10^10, y = 1/1000 .. 2, thickness = 3, discont = true);

restart;
#Taking Alejandro Jakubi's example:
S:= A*cos(2*x)*sin(3*z) + B*cos(7*x)*sin(4*z) + C*sin(3*z)+K*sin(4*z)+cos(7*x);

#Using frontend:
eval(frontend(coeff,[S,sin(3*z)]),{cos=0,sin=0});
eval(frontend(coeff,[S,cos(7*x)]),{cos=0,sin=0});
########
Edited: I have kept my first version of a procedure for this task below, but here is a much shorter one using coeffs and no freezing:

p:=proc(S::`+`) local cs,T;
   cs:=[coeffs(S,indets(S,specfunc(anything,{sin,cos})),'T')];
   select(type,`[]`~([T],cs),[specfunc(anything,{sin,cos}),anything])
end proc;
p(S);

############
Original version:

Making a procedure to find all. This is in parts like my answer to
http://www.mapleprimes.com/questions/142564-Extracting-Coefficients-From-A-Fourier-Series

p:=proc(S::`+`) local L,Lf,cs,ff,ffcs;
   L:=convert(S,list);
   Lf:=evalindets(L,specfunc(anything,{sin,cos}),freeze);
   cs:=eval(L,{sin=1,cos=1});
   ff:=Lf/~cs;
   ffcs:=`[]`~(ff,cs);
   thaw(remove(type,ffcs,[`*`,anything]))
end proc;
   
p(S);

You could try complexplot:
u:=exp(-I*f(x)*x)/sqrt(f(x));
u1:=eval(u,f(x)=sin(x)^2);
with(plots):
eps:=.1:
complexplot(eval(u,f(x)=sin(x)^2),x=eps..Pi-eps);
animate(complexplot,[u1,x=eps..a],a=eps..Pi-eps);

You may consider dividing by z-1, since z = 1 is always a root. So let f1:=f/(z-1);

I tried with Digits = 15 and parameters lambda = 25.5904477612, p = 0.0484865622, n= 27.

Using f,  26 roots were found: One was missing.
Using f1,  also 26 roots were found. So we got them all.
I doubt if this device guarantees success always, but it might help.