Use command residue. For example

residue(1/z, z= 0);

     1

The problem is that you are using m in two different ways. You are using it as an index to the middle add, and you are using it as a procedure. Try this:

t:= exp(2*Pi*I/11);
Mij:= (i,j)-> M[(i mod 11)+1, (j mod 11)+1] ;  
mu:= proc(i,j)
local n,m,k;
     add(add(add(a[k]*a[m]*a[n]*t^m*Mij(i+k-m,j+n-m), n= 0..10), m= 0..10), k= 0..10)
end proc;

Your "New problem" is caused by you trying to return an expression sequence. There is no facility for the return value of a procedure produced by unapply to be an expression sequence. There are two things you can do: (1) Make the return value a list, or (2) Make the return value a list and compose the result with op to convert it back to an expression sequence.

To do (1):

f:= unapply(
      [seq(product(t[j]^'Q[j,i]',j=1..RowDimension(Q)),i=1..ColumnDimension(Q))]
    , seq(t[i],i=1..RowDimension(Q))
);

To do (2):

f:= op@unapply(
      [seq(product(t[j]^'Q[j,i]',j=1..RowDimension(Q)),i=1..ColumnDimension(Q))]
    , seq(t[i],i=1..RowDimension(Q))
);

The solve command should contain variables u and v, not %u and %v.

You can do it like this:

for i in seq(0..100, 0.2), seq(100..0, -0.2) do ... end do;

I will assume that the backslash in your expression should have been a forward slash---that it represents ordinary division. I'll plot a unit sphere in spherical coordinates. So, I am also assuming that you want x and y to range from -1 to 1.

f:= (x,y)-> (5*x^2*y^2-6*x^4)/((1-y^2)*(x^2+y^2)-y^2*(1-x^2-y^2));

plot3d(
     1, theta= 0..2*Pi, phi= 0..Pi,
     color= f(cos(theta)*sin(phi), sin(theta)*sin(phi)),
     coords= spherical, style= patchnogrid, labels= [x,y,z]
);

Make it a single command:

eval(y, isolate(x, D(a)));

Do you want a uniform random selection from all 6-tuples of nonnegative integers whose sum is 8? It can be done like this:

C:= combinat:-composition(8+6, 6):
C[(rand() mod nops(C))+1] -~ 1;

If you want to make such a random selection repeatedly, the above can be made more efficient. Let me know.

Your attempt to trap division-by-zero errors went awry. You set it up so that division by zero returns infinity, but the plots:-circle command doesn't know what to do with infinity as an argument. Your problem can solved by simply filtering out the values that lead to division by zero; there is no need to trap those errors. 

To fix the problem, simply replace the line

d := seq(circle([1, 1/i], 1/i, color = blue), i = -20 .. 20, .1);

with

d:= seq(`if`(i=0, NULL, circle([1, 1/i], 1/i, color= blue)), i= -20 .. 20, .1):

The principles for solving the system of piecewise ODEs are the same as the simplified example above. It is done like this:

Sys1:= EQ||(1..3), op(2, EQ4), op(2, EQ5):
Sys2:= EQ||(1..3), op(4, EQ4), op(4, EQ5):
ICs1:= seq(q||k(0)=0, k= 1..5), seq(D(q||k)(0)=0, k= 1..3):
Sol1:= dsolve({Sys1, ICs1}, numeric, maxfun=0, range= 0..10):
ICs2:= eval(
     [seq(q||k(10) = q||k(t), k= 1..5),
      seq(D(q||k)(10) = diff(q||k(t), t), k= 1..3)
     ], Sol1(10)
)[]:
Sol2:= dsolve({Sys2, ICs2}, numeric, maxfun= 0):
P1:= plots:-odeplot(Sol1, [seq([t, q||k(t)], k= 1..5)], t= 0..10):
P2:= plots:-odeplot(Sol2, [seq([t, q||k(t)], k= 1..5)], t= 10..20):
plots:-display([P1,P2]);

The solution to this ODE becomes singular before t = 0.03. However, I will show how to work with an ODE like this using the function f(x,t) from your earlier post. That one doesn't go singular until about t = 15.

Individual variables can be cleared with unassign.

This may not be the most efficient way, but I think that it is robust and that it handles all the special cases.

power:= (term,x)-> `if`(term=0, undefined, expand(diff(term,x)*x/term));

Use piecewise:

f:= t-> piecewise(t < 10, 1-t, t):

f(t);

You have not defined the function for t <= 0 or t = 10. The above will of course apply the upper branch 1-t to t <= 0 and the lower branch to t = 10.

Try this, if you haven't already:

for T from 0 to 15 by 0.1 do
     your script
end do;

If that doesn't work, then you'll need to post your code.