Let me address 1. Assume that for all roots $\alpha$, there exists $t\in T(k)$ with $\alpha(t)\neq 1$. I believe this implies that $(N_{G(k)}(T(k)))/T(k)=W$. This condition is true whenever $|k|>4$, as for each $\alpha$ there is a corresponding $SL_2\to G$ and the restriction of $\alpha$ to the image of the torus in $SL_2$ is of the form $\alpha(u)=u^2$.

Consider the adjoint representation of $G$. Since $G$ is reductive, the kernel of the adjoint representation is the centre of $G$. So we lose no relevant information by only considering the adjoint representation. Let $g\in N_{G(k)}(T(k))$. Let $\alpha$ be a positive root and let $t\in T(k)$ be such that $\alpha(t)\neq 1$. Let $s=g^{-1}tg$. Note that $s\in T(k)$.

Think of $g$ as a linear operator from $\mathfrak{g}$ to itself. Writing $\mathfrak{g}=\mathfrak{h}\oplus (\oplus_{\alpha} g_{\alpha})$, let $A$ be the component of the matrix representing $g$ that sends $g_{\alpha}$ to $\mathfrak{h}$. Let $v\in \mathfrak{g}_{\alpha}$ be nonzero. Then $Av=sAv=Atv=\alpha(t)Av$ which implies that $A=0$.

This argument and a similar one in the other direction shows that the summands $\mathfrak{h}$ and $(\oplus_{\alpha} g_{\alpha})$ are $g$-invariant. Therefore $g\in N_G(\mathfrak{h})$ (the algebraic group).

Here I want to claim that $N_G(\mathfrak{h})=N_G(T)$. But Lie algebras scare me in positive characteristic and I'm not seeing it right now, hopefully someone will come along in the comments and put me out of my misery. Certainly there is an inclusion $N_G(T)\subset N_G(\mathfrak{h})$ and I recall hearing somewhere that $N_G(T)$ was a maximal Zariski-closed subgroup of $G$ which would do the job, alas I don't know a reference or proof for this.

Assuming that last part is correct, it finishes the job. And if we want to go in the other direction, I think the minimal Levi corresponding to any $\alpha$ for which no $t$ exists with $\alpha(t)\neq 1$ will lie in the normaliser but I didn't dot my i's and cross my t's when thinking about it.

Hopefully we have now an answer whenever $k\neq \mathbb{F}_3$ and a criterion that can be checked on each group for $k=\mathbb{F}_3$.

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