If there exist Moment-Generating Function of $X$, $k$-th moment $m_k$ is
$$m_k = \sum_{0\le l \le k}\dfrac{k!}{(k-2l)!l!}2^k, \;k=0,1,2... \; \text{and }l\in\Bbb N$$
Then I need to find Probability function of $X$
I had derived $m_1 = 2$, $m_2 = 12$, but these are not helpful.
Any hint?
Replacing the typo $0\leqslant\ell\leqslant k$ by the correct condition $0\leqslant2\ell\leqslant k$ in the summation that defines each $m_k$, one gets $$ E(e^{tX}) = \sum_{k\geqslant0}m_k\frac{t^k}{k!} = \sum_{k\geqslant2\ell\geqslant0}\frac{(2t)^k}{(k-2\ell)!\ell!} = \sum_{\ell\geqslant0}\frac{((2t)^2)^\ell}{\ell!}\sum_{k\geqslant2\ell}\frac{(2t)^{k-2\ell}}{(k-2\ell)!} = e^{(2t)^2}\cdot e^{2t} $$ which you should be able to recognize as the MGF of a well known distribution.