If you use the fact that $ (\ln(x!))'= \psi(x+1)$, where $\psi(x)$ is the digamma function, and l'hopital's rule, then the first limit can be evaluated directly
$$ \lim_{n\to\infty}\frac{3\cdot \sqrt{n}}{\log(n!)}=\lim_{n \to \infty} \frac{3}{2}\frac{1}{\sqrt{n}\psi(n+1)}=0 \,,$$
where the fact that $$ \lim_{n\to \infty} \psi(n+1)=\infty \,, $$
has been used.
For the second limit, recalling the asymptotic of $\ln(n!)$
$$\ln(n!)=\sum_{k=1}^{n}\ln(k)\sim \int_{1}^{n} \ln(x)dx \sim n\ln(n)-n+1, $$
we have
$$ \frac{\log(n!)}{\log(n)^{\log(n)}}\sim \frac{n\ln(n)-n+1}{\log(n)^{\log(n)}} \,.$$
Making the change of variables $m=\ln(n)$ yields
$$ \frac{\log(n!)}{\log(n)^{\log(n)}}\sim \frac{n\ln(n)-n+1}{\log(n)^{\log(n)}}=\frac{me^m-e^m+1}{m^m} \rightarrow 0 $$
as $m\to \infty,$ since $m^m>me^m \,\,\, \forall m>4. $
Note: we can use $\ln(n!)\sim n\ln(n)-n+1 $ to prove the first limit goes to $0$.