I agree. I'm happy for the question to be closed or migrated (the formulation by the author is also strange). Just thought it would be good to give some reference and finish the question.

Essentially yes. See Theorem 24.17 in the book "Linear Algebraic Groups and Finite Groups of Lie Type" by Malle and Testerman. There are some exceptions for very small groups, generally when $q = 2$, but it's a small list.

@YCor It's the ATLAS notation for a split extension.

For the case where W is a real reflection group and W' is a parabolic subgroup then the structure of the normaliser was completely determined in a paper of Howlett. See "Normalisers of parabolic subgroups of reflection groups", J. London Math. Soc. (2), vol 21 (1980).

@DanielJuteau Thanks! Sorry about that, apologies to Serre and France. I hope you'll except my apology on their behalf ;).

It's very hard to understand what's going on just by reading Digne--Michel, I wouldn't advise it. You might want to look at my answer to this question mathoverflow.net/questions/203602/…. Good luck.

Look at Example 4.3.3 in Geck's "An Introduction to Algebraic Geometry and Algebraic Groups" and the results around there.

I think you should really read Chapter 4 of Geck's book. It has all the answers to your questions. The answer here is no. There always exists an $F$-stable maximal torus and Borel subgroup $T_0\leqslant B_0$ in $G$. If you take $T = gT_0g^{-1}$ with $g^{-1}F(g) = n \in N_G(T_0)$ then this is always $F$-stable. For convenience, assume $F$ acts trivially on the Weyl group $N_G(T_0)/T_0$. Then the maximal torus $T$ is contained in an $F$-stable Borel subgroup if and only if $n \in T_0$. Hence there are lots of $F$-stable maximal tori that are not contained in $F$-stable Borels.

I'll leave the comment there but it's obviously wrong. If the element is regular then the Lie algebra of the derived subgroup is $\oplus_{i \geqslant 3}\mathfrak{g}_i$ and $\oplus_{i \geqslant 2}\mathfrak{g}_i$ is the Lie algebra of the whole radical.

Let $\mathfrak{u}$ be the sum $\oplus_{i \geqslant 2} \mathfrak{g}_i$ then I'm pretty sure this is the Lie algebra of the derived subgroup of the unipotent radical of $P$ (it certainly contains it). With some mild restrictions we know the following: $\mathfrak{c_g}(x) \subseteq \mathfrak{p}$, $[x,\mathfrak{p}] = \mathfrak{u}$ and the orbit $(\mathrm{Ad} P)(x)$ is dense in $\mathfrak{u}$. If one can characterise $\mathfrak{u}$ as above then these statements don't involve the grading, i.e., the $\mathfrak{sl}_2$-triple. The last one might be enough to pin down $P$ but I'm not sure.

I edited my answer, hopefully this is now complete. Your idea of using the closure of the fundamental Weyl chamber was the right call.