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.

Now the dictionary appears as "Definition".

Looking for ?Definition  the first definitions appear in the search list, but not all, probably because the length of the list is limited.
But searching e.g. ?Definition,taylor,  the  definitions related to taylor appear.
In my opinion, the organization and the seach system used to be better in earlier versions.