How to order the functions based on complexity $2^n$ vs $n^{\log n}$?
Firstly, your plot is wrong. See this:
https://www.wolframalpha.com/input/?i=plot%5B+y%3D2%5Ex,+y%3Dx%5Elog_2(x)%5D

Take the base 2 logarithm of them:
$\log_2 (2^n) = n$, $\log_2 (n^{\log n}) = \log n \cdot \log_2 (n)$.

As you can see, $n$ is larger than $\log n \cdot \log_2 (n)$; therefore $2^n = \omega(n^{\log n})$.

So, the answer to your first question is no. I also show you an easy proof sketch.