I suggest you apply these explanations about the $o$ and $O$ notations to the setting of the $\Omega$ notation. The take home message is that $\Omega(f(n))$ is in fact a class of sequences $(g(n))$ and not one particular sequence $(g(n))$. For example, if $g(n)=\Omega(f(n))$ (something which, in full rigor, should be denoted $g\in\Omega(f)$), then $ug(n)=\Omega(f(n))$ for every nonzero $u$.
A trivial remark is that $(f(n))$ always belongs to $\Omega(f(n))$ but many other sequences $(g(n))$ do as well (in general). In any case, since $f(n)=n^2\log(n)$ is a sequence you admit as a reference, your second case is solved by $n^2\log(n)=\Omega(n^2\log(n))$.
As regards your first example, defining $f(n)=\log(n!)$, an easy induction(1) yields $f(n)\geqslant\frac12n\log(n)$ for every $n\geqslant1$ hence $f(n)=\Omega(n\log(n))$.
(Or one can use the weak version of Stirling's approximation $\log(n!)=n\log(n)+o(n\log(n))$ and the general fact that $g(n)=f(n)+o(f(n))$ implies $g(n)=\Omega(f(n))$.)
Note
(1) If $f(n)=\log(n!)$, the induction step $f(n+1)-f(n)\geqslant\frac12\log(n+1)-\frac12\log(n)$ is equivalent to the inequality $n\log(1+1/n)\leqslant\log(n+1)$. Since $\log(1+x)\leqslant x$ for every $x$, the LHS is at most $1$ and the RHS is at least $\log(3)>1$ for every $n\geqslant2$. Since $f(1)=0$ and $f(2)=\log(2)=\frac12(2\log(2))$, this proves that the inequality $f(n)\geqslant\frac12n\log(n)$ holds for every $n\geqslant1$.