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.

restart;
f1 := x1 - x1*sin(x1 + 5*x2) - x2*cos(5*x1 - x2);
f2 := x2 - x2*sin(5*x1 - 3*x2) + x1*cos(3*x1 + 5*x2);
ds := {d1 = 3/13, d2 = 2/13};
rgs := subs(ds, [seq(seq([x1 = 2*Pi*i*d1 .. 2*Pi*(i + 1)*d1, x2 = 2*Pi*j*d2 .. 2*Pi*(j + 1)*d2], i=(floor(subs(ds, -1/d1)) .. ceil(subs(ds, 1/d1))-1)), j = (floor(subs(ds, -1/d2)) .. ceil(subs(ds, 1/d2)) - 1))]):
t0 := time():
Digits := 32:
S := NULL:
for r in rgs do
#print(r); 
S0 := NULL;
for i from 1 to 20 do
q := fsolve({f1, f2}, {x1, x2}, avoid = {S0}, op(r)):
if type(q, specfunc(anything, fsolve)) then break else S0 := S0, q fi; od:
S := S, S0;
od:
t1 := time():
print(t1 - t0);
print(nops([S]));
C := plots[pointplot](map(subs, [S], [x1, x2]), symbol = solidcircle, color = violet):
Digits := 12:
plots[display](C, seq(plots[implicitplot](L[1], x1 = -2*Pi .. 2*Pi, x2 = -2*Pi .. 2*Pi, grid = [300, 300], color = L[2]), L=[[f1, red], [f2, blue]]), size = [1200, 1200], scaling = constrained);

Also replace color specification in plot C: violet for COLOUR(RGB, .723921568, .673725492, .723921568). It looks better.

Notice that x1=0 and x2=0 cases are not in the solution set. That is probably because these are boundary cases in the relevant sub ranges. Easily solved using solve.

Moderator edit: Here's the plot, showing 190 solutions.

map2(map, convert, A, D);

with(PolynomialIdeals);
R := PolynomialRing([x, y, z, u, v], 'lex');
P := PolynomialIdeal([x^7*v, y^8*v, z^9*v, u*v]);
decomposition := PrimaryDecomposition(P);
htP := nops([decomposition]);

f := x -> x^3:
g := x -> x^(1/3):

#For complex domain these answers are correct:   
f(g(x));
g(f(x));

#For real domain these answers are correct:
A := [x < 0, x >= 0]:
f(g(x)) assuming x, real;
proc(E) global A; local a; local S := proc(E, a) a, simplify(E) assuming a end; 'piecewise'(seq(S(E, a), a = A)); end(g(f(x)));
F2 := %:

plot([Re, Im](F2), x = -2 .. 2, color = [red, blue], axes = framed);
plot([Re, Im](g(f(x))), x = -2 .. 2, color = [red, blue], axes = framed);

After simplifying for real t, res appears as:

-abs(t)^3*csgn(2*sqrt(2)+3*ln(t))+t^3;

If You know the final form then you can demonstrate with this:

R := RootOf(4*_Z^2+(4*RootOf(60*_Z^3-60*_Z^2+15*_Z-1)-4)*_Z+4*RootOf(60*_Z^3-60*_Z^2+15*_Z-1)^2-4*RootOf(60*_Z^3-60*_Z^2+15*_Z-1)+1);
final := 1/3+(1/3)*cos((1/3)*arctan(3/4));

Finalform := (1/3)*cos(alpha)+1/3;
(`@`(combine, factors, evala, Norm))(Finalform-R)[2, 1, 1];
subs(solve(%, {alpha}), Finalform);
evalf(Testzero(evalf(%-final)), 20);

#or more precisely:

Finalform := (1/3)*cos(arctan(alpha)/3)+1/3;
(`@`(combine, factors, evala, Norm))(Finalform-R)[2, 1, 1];
subs(solve(%, {alpha}), Finalform);
evalf(Testzero(evalf(%-final)), 20);

Also additional assumptions should use command additionally not assume. Using assume again on the same variable reinitializes all assumptions on it (and related variables I believe).

ode := diff(y(x), x) = sin(y(x))+1;
ic := y(0) = Pi;
Order := 40;
sol := dsolve([ode, ic], y(x), series);
ser1 := gfun[seriestorec](subs(sol, y(x)), a(n));
diffeq1 := gfun[rectodiffeq](ser1[1], a(n), y(x));
dsol1 := dsolve(diffeq1, y(x));

This should help:

indets(cS(a)(b)(c), specfunc(anything, cS));

Maple does not resolve types in the 0th operand.