time complexity – I want to show, without taking a limit, that $2^sqrt{2logn} in Ω(log^2n)$ and $2^sqrt{2logn} in O(sqrt{2}^{logn})$

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.