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;

add8 := (x,y) -> convert( convert(x,decimal,octal) + convert(y,decimal,octal),  octal,decimal );
add8(275,575);  # 1072

(The addition must be done in base 10).

Note that lists are added componentwise.

Here is a general but not very efficient procedure.
Of course it would be much better to solve the equation only once, but in this case, the solutions must be verified and the special cases must be considered.

inteq:=proc(f,x,vpar)
local vab,ab,s,T,r,n:=0;
indets(f,name) minus {x}; vab:=[op(%)];
T:=combinat[cartprod]([ [seq(vpar)] $ nops(vab) ]);
while not T[finished] do
   ab:=T[nextvalue]();
   s:=solve(eval(f, vab=~ab),x);
   if s=x then print(vab=ab,"identity!") fi;
   s:=select(type,[s],integer);
   if s<>[] then n:=n+1; r[n]:=[ab,s];DocumentTools[Tabulate]([n]) fi;
   #if s<>[] then n:=n+1; print(n,[vab=ab,x=s]) fi
od;
n,vab,x,eval(r);
end:

inteq(a*x + b = sqrt(c*x + d),x,-2..2);

            [a, b, c, d] = [0, 0, 0, 0], "identity!"
            [a, b, c, d] = [0, 1, 0, 1], "identity!"

                     176, [a, b, c, d], x,

Factorizing f7 ==>

p1-t1  or  p2-t2  or  p3-t3  is an integer multiple of Pi.

For p1-t1 = 0 (or Pi), from f1 and f2 ==> two spirals, namely

plots[spacecurve]([2.350000000*cos(t1), 2.350000000*sin(t1), t1], t1=-Pi..Pi);
plots[spacecurve]([-.350000000*cos(t1), -.350000000*sin(t1), t1], t1=-Pi..Pi);

For p2-t2 = 0 (or Pi) you get two spirals in (x,y,t2) etc.  If you want here (x,y,t1), t2 must be expressed wrt t1.

Note that the (x,y) projection is contained in 6 circles.

with(combinat, cartprod):
n:=2:
A:=Array((1..3) $ n):

T:=cartprod([[1,2,3]$n]):
while not T[finished] do
i:=T[nextvalue]();
A[i[]]:=i;  # or ... whatever
end do;

If you have Maple >= 2015, you may use the size option:

plot( max(sin(x),0)^2*sin(16*x), x=0..8*Pi, numpoints=10000,size=[1200,300]);