Proving limit with log⁡(n!) I am trying to calculate the following limits, but I dont

You can easily show that ${\displaystyle 2^n\le n!\le n^n\text{ for }n\ge 4}$ The first inequality is a very standard induction proof, and the second inequality is straight-forward (you're comparing ${\displaystyle 1\times 2\times \cdots \times n\text{ with }n\times n\times \cdots \times n).}$
From there, since ${\displaystyle f\left(n\right)=\log n}$ is an increasing function, you have that
${\displaystyle n\log \left(2\right)\le \log (n!)\le n\log \left(n\right)}$
This tells you basically everything you will need. For example, for the first one:
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{3\sqrt{n}}{n\log n}\le \underset{n\to \mathrm{\infty }}{lim}\frac{3\sqrt{n}}{\log (n!)}\le \underset{n\to \mathrm{\infty }}{lim}\frac{3\sqrt{n}}{n\log \left(2\right)}}$

Stirlings

For the first one, use that $\log (n!)\sim n\log$ n as $n\to \mathrm{\infty }$ to conclude that $\underset{n\to \mathrm{\infty }}{lim}\frac{3\cdot \sqrt{n}}{\log (n!)}=\underset{n\to \mathrm{\infty }}{lim}\frac{3\cdot \sqrt{n}}{n\log n}=\underset{n\to \mathrm{\infty }}{lim}\frac{3}{\sqrt{n}\log n}=0$