t:= `Hello Bob`;
cat(cat(t,` `)$5);

It is generally impossible to generate all the elements of the group GL(n,q) because its order is huge.
Maple knows this order (a classical formula).

with(GroupTheory):
GroupOrder(GL(4,2)); #your group
      20160
GroupOrder(GL(5,2));
      9999360
GroupOrder(GL(4,3));
      24261120
GroupOrder(GL(8,2));  # !!!
      5348063769211699200
GroupOrder(GL(n,q));  #even symbolically!
       

For symbolic result in (-Pi,Pi]

angle + 2*Pi*floor(1/2-angle/(2*Pi));

Reduce(X,X,T) is useless, it will produce only zeros.

If you want to see if a polynomial p in X is dependent of the other ones, you should call Reduce versus the rest of the polynomials i.e.

Reduce(p, Xp, T)
where
Xp := convert( convert(X,set) minus {p}, list );
Actually you should use a Groebner basis for Xp, otherwise the reduction could be incomplete (so, irrelevant).

In many situations Maple gives "generic" solutions.
For example, solve(a*x-2*a,x)  ==> x=2  (ignoring the case a=0).
However,  solve(a*x-2*a);  ==> {a = a, x = 2}, {a = 0, x = x}.
Your system has many equations of the form  a*b+c*d = 0.
solve([a*b+c*d]); ==> {a = -c*d/b, b = b, c = c, d = d},  so the solutions having b=0 are lost.

In such situation it is safer to use the Groebner package but unfortunately the system is too big for this.

They are the same, because in Sample(X,n), X can be a distribution (as in your first case) or a random variable (the second case).

Maple has difficulties because the integrand is discontinuous, symbolic and the zeros in [0,2*Pi] of
h := a*cos(t+b)+sin(t+c);
cannot be found easily in terms of a,b,c.

If you want the integral for numeric values of a,b,c, use numerical integration:
int(  eval(signum(h),[a=1,b=2,c=3]), t=0..2*Pi, numeric);

But a little math helps to see that the integral is always 0 for real a,b,c.
In fact, h can be written as
h = c*cos(t) + d*sin(t), or
h= c*sin(t+d);
Now, h is 2Pi-periodic, so
int(signum(c*sin(t+d)), t=0..2*Pi) = int(signum(c*sin(t)), t=0..2*Pi) = signum(c)* int(signum(sin(t)), t=0..2*Pi)=0.

 Edit. corrected typo: abs to signum

with(Bits):
bits:=Join([1,0,0,0,0,0,0,0]):
text:=Join([1,1,1,1,0,0,1,1]):
Xor(bits,text):
Split(%,'bits'=8);  #flip positions
add(%);  # number of flips
                    [0, 1, 1, 1, 0, 0, 1, 1]
                               5

You probably want a list.

m:=Matrix([[0,1,0,1,0,1,0]]);

 convert(m,list);
                     [0, 1, 0, 1, 0, 1, 0]

T := 2*cos(4*w):
E := sin(3*w):
plot( [E,T, w=0..2*Pi]);

eq1a:=subs(rhs(eq1)=freeze(rhs(eq1)),eq1):
eq2a:=subs(rhs(eq2)=freeze(rhs(eq2)),eq2):
solve({eq1a,eq2a},{a,e}):
thaw(%);

Maple cannot obtain a rational parametrization for curve(alpha,beta)=0, because the genus is not 0.

But, being quadratic in alpha, a parametrization can be easily obtained by solving wrt alpha.

Let's take a simpler example, to see easier the results.

ec:= (u,v) -> u^2-u+3*v^3-v-2:  # the curve is ec(u,v)=0
# choose a point on it
u0:=0.6:
v0:=fsolve(ec(u0,v),v);

vv:=v0+z^2:   # choose v as you wish
solve(ec(u,v),u);   # find parametrization for u  (exact)

eval(%[1],v=vv);   # choose the desired branch
uu:=series(%,z,8):  # obtain the approx parametrization
u=uu,v=vv;   # Parametrization near (u0,v0)

Here is a very simple one. Unfortunately it's slow for n>11.
But it is very fast for n=1  and n=2  (mod 4)    :-)

LangfordSeq:=proc(n::posint)
local A,B,C,C1,z,i:=irem(n,4),j,s,S, V:=Vector(2*n);
if i=1 or i=2 then return "Solutions exist only for n = 0 or 3 (mod 4)" fi;
A:=seq(add(z[i,j],j=1..2*n)=1, i=1..2*n);
B:=seq(add(z[i,j],i=1..2*n)=1, j=1..2*n);
C:= seq(seq(z[i,j]=z[i+n,i+j+1], j=1..2*n-i-1), i=1..n);
C1:= seq(seq(z[i,j]=0, j=2*n-i..2*n), i=1..n);
s:=Optimization:-LPSolve(0,{A,B,C,C1},assume=binary);
S:=eval(Matrix(2*n,symbol=z), s[2]);
for i to n do for j to 2*n do
  if S[i,j]=1 then V[j]:=i; V[j+i+1]:=i; fi od;od:
convert(V,list);
end:

LangfordSeq(7);

In such problems maths always beats a CAS.

f:=x -> x^4+c*x^2+x^3+d*x-c-1

==> diff(f(x),x,x)  = 12*x^2+6*x+2*c  >= 0  for  c>= 3/8.

So, f in convex in x for c>= 3/8 ==>

g(c,d) = max(f(0), f(1)) = max(-1-c, 1+d) ,  for  c>= 3/8.

So, g(c, -c-2) = -c-1, (c>= 3/8)
and
limit(g(c, -c-2), c=infinity) = -infinity

==> g is not bounded below.

Edit. g(c,d) here represents max f(x) for x in [0,1] instead of [-1,1] as was asked.
For the [-1,1] version see the comment below.

kekuncirahsia:=proc(n::posint) local r:=irem(n,9); `if`(r=0,9,r) end;