Denesting the two numbers $$\sqrt{1\pm\frac{\sqrt2}{2}}$$ should be straightforward—supposed it is at all possible?
When following the ansatz $$\sqrt{1\pm\frac{\sqrt2}{2}}\;\overset{!}{=}\;p+q\sqrt2$$ one is led to solve $$p^2+2q^2=1\quad\text{and}\quad pq=\pm\frac14$$ $\ldots$ and the cat catches its tail since $$p^2\;=\;\frac12\left(1\pm\frac{\sqrt2}{2}\;\right)$$ and similar for $q^2$.
I came across this after diagonalising a positive-definite matrix whose positive square root is needed in a proof somewhere else.
If you just want to decide if it's possible, that's easy: it would mean there is a rational solution of $8p^4-8p^2+1=0$. If $p=a/b$ with coprime integer $a,b$, it's known that $a$ must be a divisor of $1$, and $b$ a divisor of $8$, so it's a rather easy to check all cases.