@prithatiti

For shorter process, I think you can do what arjun sir did but problem is how much big 'n' can be. If you get $\lim_{n \rightarrow \infty } \frac{f(n)}{g(n)} = c$ where $c>0$ and limit exists then you can easily say that $f(n) = \Theta (g(n))$. So, it should be both $\mathcal{O}(g(n))$ and $\Omega(g(n)).$ Sometimes, manipulation in functions might work like this to solve these type of questions. So, if you have 2 functions like $2^{\lg 2n^{10}}$ and $n^{10}$. With some manipulation, you have changed $n^{10}$ to $2^{\lg n^{10}}$ then we can say, both functions have same growth rate and can be written in theta notation to each other.

Here, options (c), (d) and (e) can be easily eliminated.

$\log^2 n \approx (\log m^m)^2 = (m\log m)^2 = m^2 \log^2m$. Here, $m \in \mathcal{O}(m^2log^2m)$ but $m \notin \Omega(m^2log^2m)$.

Same thing you can do for function $\log^{1.5} n.$