Your problem has undecifered trig identities, hence problems. Consider:

eqs:=[_C1+_C2 = 0, _C1*exp(3^(1/2)*((cos(1/6*Pi*3^(1/2))-1)*(cos(1/6*Pi*3^(1/2))+1))^(1/2)/(cos(1/6*Pi*3^(1/2))-1)^(1/2)/(cos(1/6*Pi*3^(1/2))+1)^(1/2)*ln(cos(1/6*Pi*3^(1/2))+(cos(1/6*Pi*3^(1/2))^2-1)^(1/2)))+_C2*exp(-3^(1/2)*((cos(1/6*Pi*3^(1/2))-1)*(cos(1/6*Pi*3^(1/2))+1))^(1/2)/(cos(1/6*Pi*3^(1/2))-1)^(1/2)/(cos(1/6*Pi*3^(1/2))+1)^(1/2)*ln(cos(1/6*Pi*3^(1/2))+(cos(1/6*Pi*3^(1/2))^2-1)^(1/2))) = 4];

convert(combine(simplify(evalc(radnormal(convert(eqs, tan))))), tan);

#[_C1+_C2 = 0, I*_C1-I*_C2 = 4]

Edit:
also works with simplier:
combine(simplify(evalc(eqs)));
evalc(simplify(expand(eqs)));

 

If you mean order preserving, distinct selection from n characters:

F := (S :: string, q :: nonnegint) -> ((A, q :: nonnegint) -> map(op, [op](combinat[randcomb](nops(A), q)), A))([seq(S)], q);
F("abcdefghijklmnopq", 8);
F("a1=#@B", 3);

However, repeating this application k times does not guarantee these elements would be different selections.

eval('match(2*y(3)^4+5=k1*y(k2)^k3+k4,Y, la)', [y(3)=Y,y(k2)=Y]);
la;
 

numArray := [17, 18, 19];
symArray := [x, y, z];
kronArray := map(proc(i) local j; symArray[i] = symArray[1]^mul(numArray[j], j = 1 .. i - 1) end, [$ (1 .. 3)]);

a = 1/y, b = x/(x+y-1);
solve({%}, {x, y});
{x = (y-1)^2, 1 < y, y < 5/4};
subs(`%%`, %);
solve(%);

Also, did you look at this result:
solve({a = 1/y, b = x/(x+y-1), x = (y-1)^2, 1 < y, y < 5/4}, {x, y}, parameters = {a, b});

Try:

printlevel := 3;
for x in [ -1, 1 ] do for y in [ -1, 1 ] do 'x*y'=x*y; end do; end do;

printlevel := 2;
for x from 1 to 2 do for y from 1 to 2 do 'x*y'=x*y; end do; end do;

Include sol:=radnormal(sol); in your code before ode test and everything turns fine. That radical of yours is very poorly presented. Maple will not waste any effort doing this for you. In general fidling radicals is an expensive and unproductive affair.

That last one should have been:

eqs := [a = b, c = d, e = f]:
map2(map, `^`, eqs, 2); #or replace map2 with map[2]

Is this the answer:
[x, y] in {[-2, 2], [0, 0], [1, 1], [2, 2]};

There are 98 subcases in DNF form. All of them point to empty set.

Solving H1..H6 we get:

{-1.960600500 <= tau2, 0.2072815534e-2 <= tau1, tau1 <= 0.2365533980e-2, tau2 <= 0.3994999999e-3}

to satisfy objective 0<=H<=1.

It is as reliable as solve command, which in this example fails.

{solve}(radnormal(diff(e, x), rationalized), {x});

The problem is e represents a root of quadratic, rather than an apparent higher degree root:

e := sqrt(2)*sqrt((3*x^3 + 5*sqrt(x^2 + 8)*x^2 + 24*x + 36*sqrt(x^2 + 8))/sqrt(x^2 + 8))/8 + x/8 + (3*x^2)/(8*sqrt(x^2 + 8)) + 3/sqrt(x^2 + 8);
e2 := (1/4)*x+3*sqrt(x^2+8)*(1/4);
(evala@Norm)(e2-e);
f := lambda -> -lambda*(3*(x^2+8)^(1/2)-4*lambda+x);
radnormal(f(e));
 

The result of:
solve({x*(1-y)/y*(1-x) < t,  0 < x, x < 1, 0 < y, y < 1, t > 0}, {y}, parameters = {x, t});
is quite self explanatory. So any of these should suffice:
plt1 := t -> plot([-x*(x-1)/(-x^2+t+x), 1], x = 0 .. 1, color = [red$2], thickness=[2$2]);
plt1(0.3);
plt2 := t -> plots[inequal]([-x*(x-1)/(-x^2+t+x) < y, y < 1], x = 0 .. 1, y = 0 .. 1);
plt2(0.3);
 

Not much to say here. Would the solution solf := ln(y-1)+2*arctanh(y^(1/2))-I*Pi; be better?
evalc(simplify(Im(solf)));
plot(%, y = -4 .. 4);
Complex analisys is the Maple way.

Hope this helps (edited):

restart;
A[1]:=alpha=1-(1/2)/(1-(RootOf(16*_Z*(_Z*(2*_Z*(_Z*(8*_Z*(_Z*(_Z*(_Z*(32*_Z*(8*_Z-33)+1513)-812)-13)+267)-1469)-330)+811)+279)+345,index=2)-1/2)**2);
A[2]:=expr=(1+alpha)*sqrt(1-alpha**2)+(3+4*alpha)/12*sqrt(3-4*alpha**2)+2*(1+alpha)/3*sqrt(2*(1+alpha)*(1-2*alpha))+(1+2*alpha)/6*sqrt(2*((1-alpha)**2-3*alpha**2));
alias(beta=indets(A[1],RootOf)[1]);
F := lambda^2+(8159503/855552)*beta-(236960/1671)*beta^9+(2085340/5013)*beta^8-(13276063/40104)*beta^7+(414451/20052)*beta^6+(10073755/160416)*beta^5-(13077487/160416)*beta^4+(1471997/213888)*beta^3+(5380211/160416)*beta^2+1770613/2566656;
evalf(subs(lambda = expr, evalf(subs(A[1], A[2]), 40), F), 40);
F:=(evala@Norm)(F);
evalf(subs(lambda = expr, evalf(subs(A[1], A[2]), 40), F), 40);

Also, the file you've submitted appears corrupted, so says Maple 2017.
Sorry, my previous posting was based on a typo. This one is corrected.