Let u=sin(x) v=cos(x). We know u+v, and since (u+v)^2=1+2uv, we know uv, hence we know a basis of the symmetric functions in u,v, and we can, if we are bothered work out, u^5+v^5. That I feel is far less interesting than the knowledge about symmetric function theory...
But the OP seems to be posting all these questions as challenges, rather than questions for which he seeks the answer.
sin(x)+cos(x)=1/3 then 1+2sin(x)cos(x)=1/9= 1+sin(2x)
so sin(2x)=-8/9 then 2x=arcsin(-8/9) once you have got 'x' the rest is easy