I would like to find an asymptotic relationship between those two functions, and I don't know how, can anyone tell me a few tips or a way how to do that?
first pair:
$$(\log n)^n \quad \text{and} \quad n^{log(n)}$$ the log base is $2$.
second pair:
$$n!\ln(\ln(n)) \quad \text{and} \quad (2^n)\ln(n)$$
thanks for the help!
Here is a key lemma (which you can and should prove for yourself) that helps to resolve so many questions of this type. I'll phrase it in terms of functions of $x$ where $x\to\infty$, although it holds for other domains as well.
Lemma: Let $f,g\colon [1,\infty)\to[2,\infty)$ be functions. If $\log f(x) = o(\log g(x))$ as $x\to\infty$, then $f(x) = o(g(x))$ as $x\to\infty$.
(Important details: the lemma would be false with $o()$ replaced by $O()$—take $f(x) = x^2$ and $g(x)=x$. The lemma would also be false if the values of $f$ could go below $1$, but even if they could get arbitrarily close to $1$—take $f(x) = 1+1/x$ and $g(x)=2$.)
Note that this is also a way to prove that $f(x) = O(g(x))$, since that's a weaker statement than $f(x) = o(g(x))$; it's just important that the hypothesis has $o()$ and not $O()$ in it (as remarked above).
For your first example, the logarithms of the two functions are $n\log\log n$ and $(\log n)^2$; since the second of these is $o$-of-the-first, the lemma implies that $n^{\log n} = o( (\log n)^n )$. You should be able to work out the second one the same way, using Stirling's formula to approximate $\log(n!)$.