Maple writes about an error, you have something mixed up with parentheses. I fixed it and got a constant:

Expr:=(1/10)*exp((2/135)*sqrt(-11)*sqrt(225)*t-(2/3*I)*x+1/15)/((3/10)*exp((2/135)*sqrt(-11)*sqrt(225)*t-(2/3*I)*x+1/5));
simplify(Expr);
plot3d(%, x=-1..1, t=-1..1);

Unfortunately, your system does not have real solutions:

restart;
sys:=[(-x3*cos(x5)+(1010-x4)*sin(x5))^2/x1^2+(-x3*sin(x5)-(1010-x4)*cos(x5))^2/x2^2=1, ((-50.5-x3)*cos(x5)+(1060.5-x4)*sin(x5))^2/x1^2+((-50.5-x3)*sin(x5)-(1060.5-x4)*cos(x5))^2/x2^2=1, ((404-x3)*cos(x5)+(1313-x4)*sin(x5))^2/x1^2+((404-x3)*sin(x5)-(1313-x4)*cos(x5))^2/x2^2=1,  -2*x2^2*cos(x5)^2*x3+2020*x2^2*cos(x5)*sin(x5)-2*x2^2*cos(x5)*sin(x5)*x3-2*x1^2*sin(x5)^2*x3-2020*x1^2*sin(x5)*cos(x5)+2*x1^2*sin(x5)*cos(x5)*x4=0, x5=Pi/2]:
solve(sys);

Of course, you can easily insert the cross between two symbols directly without using any special procedures, and in two ways (in infix or prefix notation).

The first way:
`a × b`;  # Infix notation
                              

The second way:
`×`(a, b);  # Prefix notation
                             

 In the second method, arguments need not necessarily be symbols. For example, for your example, you can write:

with(LinearAlgebra):
R := Vector(3, [x, y, z]):                                           
V := Vector(3, [u, v, w]):
`R × V`=R &x V;
# or
`×`(R,V)=R &x V;

Your more complicated example:

U:=`R × V`;
`×`('R', cat(`(`, U, `)`));

with(plots):
with(plottools):
animate(spacecurve,[[sqrt(1-s^2)*cos(t),sqrt(1-s^2)*sin(t),s], t=0..2*Pi, color=red, thickness=3], s=-1..1, background=display(sphere()), frames=60, scaling=constrained, view=[-1.2..1.2,-1.2..1.2,-1.2..1.2]);

    Addition - another option: 

with(plots):
with(plottools):
f:=phi->rotate(spacecurve([cos(t),0,sin(t)], t=0..2*Pi, color=red, thickness=3), phi, [[0,0,0],[0,0,1]]):
animate(display,[f(phi)], phi=0..2*Pi, background=display(sphere()), frames=60, scaling=constrained, view=[-1.2..1.2,-1.2..1.2,-1.2..1.2], axes=normal); 

restart;
solve(sin(x)=0, allsolutions);
about(_Z1);

Addition. If you need to solve this equation at some interval, then solve the system to which you need to add  explicit  option.

Example:

solve({sin(x)=0, x>=0, x<=3*Pi}, allsolutions, explicit);
 

To get a  solution, I took an almost zero derivative  D(y)(2945)=0.00001  instead and used Preben's method from here :

restart;
ode:=diff(y(x),x,x)-0.00003019*abs(y(x))^0.337=9.542*10^(-13)*x; 
bcs:=y(0)=0,D(y)(2945)=0.00001;
ode2:=evalindets(ode,`^`,x->op(1,x)^(c*op(2,x)));
res:=dsolve({ode2,bcs},numeric,method=bvp[midrich],continuation=c,maxmesh=8192);
plots:-odeplot(res,[x,y(x)]);

Edit.
 

The first way:

restart;
alias(s[1]=sin(theta1), c[1]=cos(theta1)):
sin(theta1)*cos(theta1);

The second way:

restart;
s[1]:=sin(theta1): c[1]:=cos(theta1):
s[1]*c[1];

The third way:

restart;
subs([sin(theta1)=s[1],cos(theta1)=c[1]], sin(theta1)*cos(theta1));

Example:

evalc~([solve(z^3=1+I)]);  # All the roots of the third degree of  1+I  in a+I*b form
convert(%, radical);  # Conversion to radicals
convert~(%, polar);  # Conversion to polar 

restart;
rts := [-1.04922510287257-0.450627300642352*I, -1.04922510287257+0.450627300642352*I, 0.324928502807894-1.26649311642498*I, 0.324928502807894+1.26649311642498*I, 0.448593200129344]:
min(abs~(rts)[ ]);

                                                   0.4485932001

e2_3:= w^(1-sigma)*((f__11*sigma*Delta__1/L__1)^(-1/(sigma-1))/w)^(k-sigma+1)*k/(Delta__1*(k-sigma+1)*a__0^k);
simplify(e2_3, symbolic);

hf:=<1/(1-x1-x2)^2+1/x1^2, 1/(1-x1-x2)^2; 1/(1-x1-x2)^2,1/(1-x1-x2)^2+1/x2^2>; # Matrix
x:=<10, 20>; # Vector
subs([x1=x[1], x2=x[2]], hf);

You wrote  x:=[10, 20];  This  x  is a list not a vector
 

plot([Re,Im]~([fsolve(x^3 = 1, complex)]), style=point, color=red, symbol=solidcircle, symbolsize=20, scaling=constrained);

Explanation. In fact, in one line we combined 3 steps.

The first step is to find three roots:

L:=[fsolve(x^3 = 1, complex)];
L := [-.500000000000000-.866025403784439*I, -.500000000000000+.866025403784439*I, 1.]
 

The second step is to extract the real and imaginary parts from each root:

[Re,Im]~(L);
[[-.500000000000000, -.866025403784439], [-.500000000000000, .866025403784439], [1., 0.]]
 

The last step is the plotting of these three points by plot command.

First we solve the system if in the first equation we replace the inequality by equality. The second equation is squared (we add the condition  -2*x^2+y^2>=0  for preserving the equivalence):

solve({y = 4*x^4+4*x^2*y+1/2, (1/2)*(x-y)^2-(x-y)^4 = (-2*x^2+y^2)^2, -2*x^2+y^2>=0}, [x, y]);

                                        [[x = -1, y = -3/2], [x = 0, y = 1/2]]

As for the initial system, all solutions will be obtained if we add to the points obtained all the points of the blue curve lying in the green region:

A:=plots:-implicitplot([y = 4*x^4+4*x^2*y+1/2, (1/2)*(x-y)^2-(x-y)^4 = (-2*x^2+y^2)^2, -2*x^2+y^2=0], x=-3..3, y=-3..1, color=[red,blue, green], thickness=2, gridrefine=5):
B:=plots:-inequal({-2*x^2+y^2>=0, y<=0},x=-3..3, y=-3..1, color="LightGreen"):
plots:-display(A, B, scaling=constrained);  

What do you want to get as

is an erroneous result, which is easily verified by differentiation. On the other hand, Maple calculates everything correctly.

See  

Download ccc_new.mw

The solution in two steps:

restart;
with(IntegrationTools):
M:=Int(diff(B(x,y),x,x)*F(x,y),[x,y,z])+Int(diff(E(x,y),x,x)*H(x,y),[x,y,z])+Int(diff(E(x,y),y,y)*F(x,y),[x,y,z])+Int(diff(B(x,y),x,x)*H(x,y),[x,y,z])+Int(diff(F(x,y),x,x)*F(x,y),[x,y,z])+Int(diff(E(x,y),x)*F(x,y),[y,z])+Int(diff(E(x,y),y)*H(x,y),[y,z])+Int(diff(B(x,y),y)*F(x,y),[y,z]):
M1:=map(t->applyop(collect,1,t, [F(x,y),H(x,y)]), Combine(M));
map(Int,op([1,1],M1),[x,y,z])+map(Int,op([2,1],M1),[y,z]);  # The final result 

The code was edited.