mtaylor(sin(x+y)+y, [x=Pi, y=0], 6);

sinc:= x -> sin(x)/x:
sinc(0):=1: sinc(0.):=1.:
n:=10;  #1000 if you insist
a:=5.0; h:=a/n;
Z:=Matrix(2*n+1, (i,j) -> sinc((i-1-n)*h)^2*sinc((j-1-n)*h)^2, datatype=float[8]);

# note that sinc is even, so Z is symmetric.

You can convert using IrfanView (free)
http://www.irfanview.com/main_formats.htm

Explore(plot([(-y0-x0)*exp(-2*t)+(y0+2*x0)*exp(-t),
               -2*(-y0-x0)*exp(-2*t)-(y0+2*x0)*exp(-t), t=0..5]),
        x0=1..5., y0=1..5.);

evalb(irem(390,7)=0);
   false

Q:=[X,Y]:
Ps:=proc(f,g)
local k;
global P,Q; # not really necessary
add(diff(f,P[k])*diff(g,Q[k]) - diff(f,Q[k])*diff(g,P[k]),k=1..nops(Q))
end: 

Ps(P[1],X) ;

with(plots):with(plottools):
X:=cos(t):Y:=sin(t):
vol:=Pi*X^2*(2*Y) + 2*Pi*X^2*(1-Y)/3:
sol:=maximize(vol,t=0..Pi/2,location):
t:=eval(t, op([1,1],sol[2]));#evalf(%)*180/Pi;
p1 := display(polygon([[[-X, Y], [0, 1], [X, Y], [X, -Y], [0, -1], [-X, -Y], [-X, Y]]]), color = red, thickness = 2):
p2 := implicitplot(x^2+y^2 = 1, x = -1 .. 1, y = -1 .. 1):
display(p1, p2);
simplify([X,Y]);

#[X,Y]

It is not fsolve, it is your system which contains

[c0 = 0, ... , (...)/c0 + ... = 0, ...]

Of course it cannot be solved. Replace ":" by ";" to see what happens.

PS. Why do you use such strange constructs?
F(eta) := ...
...
F(eta) := unapply( F(eta), eta);
Sintactically it is correct but now F is an empty procedure and in its remember table F(eta) is also a procedure. It works, but why so?
 

with(Statistics):
m:=Mean(a):
v:=Variance(a):
b:=v/m: c:=m/b:
P:=DensityPlot(RandomVariable(Gamma(b,c))):
plots[display](h,P);

You may clip the plots manually instead of using view.
For example,

g:=sin(k*x):
g1:=piecewise(g<1/2, g, undefined):
plots:-animate(plot, [g1, x=0..4, view=[-6..6,-2..2]], k=1 .. 6.0);

You cannot expect acceptable results for this model: your function takes values < 1 (for any parameters) while your y-vector has almost all values > 20.

In Windows you could write a short Autoit script to "export as .mpl" (see https://www.autoitscript.com) and insert a ssystem call  in the worksheet which executes the script.

fdiff at x needs the values of the function at x + h and x - h etc and temperature is not defined for x>1.
You can use e.g.:

evalf((D[1](temperature))(0.99999, .5));
or

h:=10^(-5);
(-temperature(1-h, .5) + temperature(1, .5))/h;

restart;
a:={theta <= Pi-arccos(-3/4+(1/4)*sqrt(13+16*sqrt(2))), arccos(3/4) < theta};
assume (op(a));
getassumptions(theta);  op([1,2],%);
theta:='theta'; # forget properties

Another workaround is to use 1D input.
[Because in 2D,
  d:=(1..3)$n 
also fails, and for n=1000 there is no other option].

P.S. I wonder if the following theorem is true:
For any 2D parser P, there exists a correct 1D expression for which P fails.