To prove $2^{sqrt{2log n}} in O(sqrt{2}^{log n})$, you should note that $sqrt{2}^{log n} = left(sqrt{2}^{sqrt{log n}}right)^{sqrt{log n}}$.

To prove $2^{sqrt{2log n}} in Omega(log^2n)$, you should note that $log ^2n = 2^{2log log n}$.

It will simplify the comparisons you have to do.