It is well known that a square system has a unique solution if and only if it's main determinant is not 0.

restart;

A:= < k+3, 2, k-4, 3;

      0,   2,  -9, 5;

      0,   0, k^2+k-2, k-1 >;

B:=A[..,1..3];

V:=A[..,4];

LinearAlgebra[Determinant](B);

Sol:=[solve(%)];

We have a unique solution for any number  k  not included in this list.

Next solve separately for each number from  list  Sol:

for i from 1 to 3 do

'k'=Sol[i],  LinearAlgebra[LinearSolve](op(eval([B,V],'k'=Sol[i])), free=C);

od;

Thus for  k=-3  and  k=1  we have infinitely many solutions, for  k=-2  no solutions

You made ​​a few syntax errors, which I corrected.
Your nonlinear system has 5 parameters. Maple can not solve it symbolically for all values ​​of the parameters, but Maple can  solve this system for the specified parameters.

restart;

f := proc (E2, T, DHT, Ca, Cshbg)

local KsT, KaT, KsE2, KaE2, KsDHT, KaDHT, eq1, eq2, eq3;

KsT := 0.10e11; KaT := 4.6*0.10e6; KsE2 := 3.14*0.10e10; KaE2 := 4.21*0.10e6; KsDHT := 3*0.10e6; KaDHT := 3.5*0.10e6;

eq1 := E2 = fE2*(1+(KaE2+Ca)/(1+KaE2*fE2+KaT*fT+KaDHT*fDHT)+KsE2*Cshbg/(1+KsE2*fE2+KsT*fT+KsDHT*fDHT));

eq2 := T = fT*(1+KaT*Ca/(1+KaE2*fE2+KaT*fT+KaDHT*fDHT)+KsT*Cshbg/(1+KsE2*fE2+KsT*fT+KsDHT*fDHT));

eq3 := DHT = fDHT*(1+KaDHT*Ca/(1+KaE2*fE2+KaT*fT+KaDHT*fDHT)+KsDHT*Cshbg/(1+KsE2*fE2+KsT*fT+KsDHT*fDHT));

solve({eq1, eq2, eq3})

end proc:

Example:

f(1, 2, 3, 4, 5);

We got 10 solutions.

Procedure  UniqueColumns  returns  the unique columns of a Matrix.

UniqueColumns:=proc(A::Matrix)

local L;

uses LinearAlgebra, ListTools;

L:=[Categorize((x,y)->Equal(x,y), [seq(A[..,i],i=1..ColumnDimension(A))])];

op(map(op,select(x->is(nops(x)=1), L)))

end;

Your example:

Q := Matrix([[1,1,1,0,0,0,-2,-1], [0,0,0,1,1,1,0,-3]]);

UniqueColumns(Q);

Write your function as a procedure:

f:=proc()

your expression

end;

Example:

f:=proc()

add(args[i]^2, i=1..nargs)

end;

f(1, 2, 3);

       14

I do not understand the meaning of the problem. If you want to write it in the same loop then 

for i in  [seq(i, i=0..100, 0.2), seq(i, i=100..0, -0.2)]  do ... od;

Multiplication signs, division sign and missing parenthesis

((5*x^2*y^2)-6*x^4)/((1-y^2)*(x^2+y^2)-(y^2*(1-x^2-y^2)));

The parenthesis can be put in another place.

This is arithmetical progression.

f:=unapply(rsolve({u(n+1)=u(n)+d, u(1)=a}, u(n))/b, n);

                              f := n -> (a+d*(n+1)-2*d)/b

Example:

eval(f(100000000000), {a=25, d=25, b=5});

                                         500000000000

Look at my posts   http://www.mapleprimes.com/users/Kitonum/posts 

You can find a number of interesting applications of them.

You can use geometric transformations from  plottools  package.

Examples:

with(plottools): with(plots):

F:=curve([[-4,0],[0,3],[1/2,0], [-4,0]], thickness=3):

F1:=reflect(F, [[-1,4],[1,4]]):

F2:=translate(F, -4, 2):

F3:=rotate(F, -Pi/2, [3,0]):

T:=textplot([[-1,1,"F"], [-1,7,"F1"], [-5,3,"F2"], [4,4,"F3"]], font=[TIMES,ROMAN,20], color=blue):

display(F, F1, F2, F3, T, scaling=constrained);

Code of procedure:

ExponentOfPower:=proc(monom, var)

`if`(not has(monom,var), 0,  `if`(nops(select(has,monom,var))>1, op(2,select(has,combine(monom),var)), 1));

end:

Examples:

term1:=3*x^k*y^(k+2): term2:=3*y^(k+2): term3:=3*x*y^(k+2): term4:=3*x^k*y^(k+2)*x^2:

op(map(ExponentOfPower, [term||(1..4)], x));

                                                                        k, 0, 1, k+2

Example with Carl's procedure:

power(x^x, x);

             x*ln(x)+x

f := t->piecewise(0 < t and t < 10, 1-t, 10 < t, t);

plot([f(t), [10, t, t = -9 .. 10]], t = 0 .. 15, thickness = [2, 1], color = [red, black], linestyle = [1, 2], discont, view = [-2.5 .. 14.5, -10 .. 14.5], scaling = constrained);

term:=3*x^k*y^(k+2):

op(2, select(has, term, x)),   op(2, select(has, term, y));

                                                  k, k+2

There is no triangle with sides 136, 170, 174, in which the coordinates of all vertices are integers or rational numbers. It is well known that the area of any such triangle is rational. But the area of your triangle is irrational:

a := 136: b := 170: c := 174:

p:=(a+b+c)/2:

sqrt(p*(p-a)*(p-b)*(p-c));

                                        240*2002^(1/2)

Tools -> Options -> Precision -> Round calculation ... -> Apply to Session or Apply Globally

1) Your code is very inefficient. To specify the sequence of the objects is better to use the  seq  command. Also I do not understand the meaning of the options  coords=polar, axiscoordinates=polar, as your circles given in Cartesian coordinates.

with(plots): with(plottools):

c := seq(circle([i/(1+i), 0], 1/(1+i), color = green), i=1..20):

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

display(c, d, view = [0 .. 2, -0.5 .. 2], scaling=constrained);

2) intersect command works only with finite sets. In order to find the point of intersection of the two lines, you can use solve  command.

Sol := solve({(x-1)^2+(y-1)^2 = 1, (x-2/3)^2+y^2 = 1/9}, explicit);

                        {x = 1, y = 0}, {x = 2/5, y = 1/5}

Point1 := disk([rhs(Sol[1, 1]), rhs(Sol[1, 2])], 0.01, color = red):

Point2 := disk([rhs(Sol[2, 1]), rhs(Sol[2, 2])], 0.01, color = red):

display(L, LL, Point1, Point2, view = [0 .. 1.5, -0.5 .. 0.5], scaling = constrained);

Edited. Fixed equation of  c .