Let $G$ be a semisimple algebraic group over the complex numbers and we fix a maximal torus $T$. Let $w\in W$ be an element in the Weyl group, and let $T^{w}$ be the elements in $T$ that are fixed by $w$. Is $T^{w}$ still a torus? Is the Lie algebra of $T^{w}$ going to be $t^{w}$, the fixed points of $w$ acting on the Cartan subalgebra?
The first statement amounts to determining whether $T^w$ is connected. In the simplest case $G={\rm SL}_2$, $T$ may be identified with the multiplicative group ${\rm G}_m=\Bbb{C}^{\times}$, the Weyl group is cyclic of order 2 and the simple reflection $s$ acts by $x\mapsto x^{1}$. Therefore, its fixed point set is $\{x: x^2=1\}=\{1,1\}$, which is disconnected and hence not a torus.
The second statement is true and holds in a more general setting of an automorphism of a Lie group: If $H$ is a connected Lie group with Lie algebra ${\frak{h}}={\rm Lie}(H)$ and $\alpha: H\to H$ is an automorphism, then ${\rm Lie}(H^{\alpha})={\rm Lie}(H)^{\alpha}={\frak{h}}^{\alpha}$. This follows from the basic Lie correspondence between Lie groups and Lie algebras.
By the way, $T$ is a torus; "tori" is the plural form.

$\begingroup$ It may be just barely worth noting that the fact about Liegroups and automorphisms is very much a characteristic0 thing, that doesn't extend 'for free' to algebraic groups over imperfect fields. $\endgroup$ – LSpice Sep 5 '18 at 18:43