Your IC and BC are inconsistent.

You have 

ic := u(x, 0) = cos(x), u(0, t) = 0;

First condition says that for all x, at t=0, u=cos(x). 

The second says at x=0, for all t, u=0.

But at x=0, from the first conditions, it says u=1, since cos(0)=1 which is not zero according to second condition.

I could not understand your worksheet. But to solve this pde you could do the following

restart;
pde := diff(u(x,t),t) = k*diff(u(x,t),x$2)+h(x,t);
bc  := u(0,t) = 10,u(1,t) = 20;
ic  := u(x,0) = 60*x -50*x^2+10;
sol:=pdsolve([pde,bc,ic],u(x,t))

For specific h(x,t)

sol:=simplify(subs(h(x,tau)=0,sol))

you could see example in

https://rosettacode.org/wiki/Associative_array/Creation#Maple

"Maple tables are hashed arrays. A table can be constructed by using the table constructor."

T := table( [ (2,3) = 4, "foo" = 1, sin(x) = cos(x) ] );
          T := table(["foo" = 1, sin(x) = cos(x), (2, 3) = 4])
 
> T[2,3];
                                   4
 
> T[sin(x)];
                                 cos(x)
 
> T["foo"];
                                   1

if it for display, you could always do

a*z^``(-1)

I do not do plotting much in Maple, so this might not be the optimal solution. it seems to lose the minor ticks though.  There might be a way to also keep these there. I think the problem is that Maple needs a little bit more space, and so it added ticks by skipping every other value. that is why the odd values 5 and -5 do not show up.

R:=10:
r:=5:
plot3d( [ ( R+r(u,v)*cos(v))*sin(u),
          ( R+r(u,v)*cos(v))*cos(u),
            r*sin(v)
        ],
            u=0..2*Pi,
            v=0..2*Pi,style=patchnogrid,
            scaling=constrained,
            scaling=constrained,
coordinateview=[-15..15, -15..15,-5..5],tickmarks=[default,default, [seq(i,i=-5..5,2)]],
lightmodel=light3, viewpoint = circleleft);

Or you could increase the view to coordinateview=[-15..15, -15..15,-6..6] to give it more space? 

Maple uses last name evaluation for number of expressions. But this does not apply to things that are rtable based. This includes Matrix, Vector, and Array etc...

last name evaluations is what you are asking for. This means, when you write A:=C; and then assign some value to C, then when you use A again, you expect it to have same value you assigned to C. This is because Maple used last name evaluation to A.

But this does not work with rtables.  That is why when you print the matrix again, you did not see its value changed after you changed m

I do not know if it is possible to change this. But you can always use eval or subs

restart; 
u:=<-2+m,3+m>;
m:=5;
u;

type(u,'last_name_eval')

false

You can now use eval to force value of m to be used inside rtable

unassign('m');
u:=eval(u,m=5)

Look up last name evaluation in help for more information.

I get slighlty different result from that shown.

f:=t->862.979774+1825.011621/((1+1.54943476*e^(-677.0138344*t))^(131537/203808));
convert(f(t),rational)

Copied the integrand to Mathematica and it gives

Copied the result from Mathematica back to Maple

expr:=`(1450669*t)/
  1681 + (4459704101248*
    Hypergeometric2F1[131537/203808, 1, 335345/203808, 
          (36887*E^((342569*t)/506))/(57154 + 
        36887*E^((342569*t)/506))])/
     (1067737900495*(1 + 57154/(E^((342569*t)/506)*36887))^(131537/
       203808))`;
maple_expr:=MmaTranslator:-FromMma(expr);
lprint(maple_expr);


1450669/1681*t+4459704101248/1067737900495*hypergeom([131537/203808, 1],[335345
/203808],36887*exp(342569/506*t)/(57154+36887*exp(342569/506*t)))/(1+57154/
36887/exp(342569/506*t))^(131537/203808)

Compare to answer by Axel 

Here is plot on top of each others, which shows some difference. I am not sure which is the correct one

Axel:=2.695678478*hypergeom([1., 1., 1.645396648],[2., 2.],-1.549434760*exp(-677.0138340*t))*exp(-677.0138340*t)+.95671881+1887.991395*t;
plot([maple_expr,Axel],t=-0.01 .. 0.01)

 How to get time dependence for fixed x?

why do you need to solve this numerically? Maple can give exact analytical answer

restart;
PDE := diff(u(x,t),t)=1/10*diff(u(x,t),x,x);
IBC := u(x,0)=1, u(0,t)=0, D[1](u)(1,t)=0;
sol := pdsolve([PDE,IBC]);

To plot for fixed x, with changing t, simply set x to some value in the above, and plot for changing t.

f:=unapply(subs(infinity=20,rhs(sol)),x,t)

plot(f(1.5,t),t=0..5)

Actually Maple's result is correct.