To answer Question 1, this can be proved in a similar manner to the case of a path algebra of a quiver, where one exhibits a "standard resolution" of a general module over the path algebra.
I will denote $A = T_R(V)$. To begin, the kernel of the multiplication map $\mu \colon A \otimes_R A \to A$ is known to be isomorphic to $A \otimes_R V \otimes_R A$. For instance, see Propositions 2.5 and 2.6 of the paper Algebra extensions and nonsingularity by Cuntz and Quillen. This provides a short exact sequence of $(A,A)$-bimodules $$0 \to A \otimes_R V \otimes_R A \overset{j}{\to} A \otimes_R A \overset{\mu}{\to} A \to 0.$$ (I imagine that with enough patience, one can show that $j$ is the usual map induced by $j(1 \otimes v \otimes 1) = v \otimes 1 - 1 \otimes v$ as in the path algebra case, but this will not be necessary to compute the global dimension.) Note that the surjection $\mu$ is split as a morphism of right $A$-modules by the map $a \mapsto 1 \otimes a$. Thus, when considered as a sequence of right $A$-modules, this is a split exact sequence.
Now consider an arbitrary left $A$-module $M$. Becasuse the sequence is right-split, it remains exact after applying the functor $- \otimes_A M$, yielding the short exact sequence of left $A$-modules
$$0 \to A \otimes_R V \otimes_R M \to A \otimes_R M \to M \to 0.$$
(If $j$ above is shown to be the expected map, then this would be the precise analogue of the usual standard resolution for a quiver algebra.) Because every left $R$-module is projective (as $R$ is semisimple), and because extension of scalars $A \otimes_R -$ preserves projectivity of a module, the sequence above is in fact a projective left $A$-module resolution of $M$. So $M$ has projective dimension at most 1.
Thus $T_R(V) = A$ has left global dimension at most 1, and similarly for its right global dimension (by the symmetric argument). If $V = 0$, then in fact $T_R(V) = R$ still has global dimension 0. But if $V \neq 0$, then I suppose one can show that the left (or right) module surjection $T_R(V) \twoheadrightarrow T_R(V)/\left(\bigoplus_{k \geq 1} V^{\otimes k}\right) \cong R$ is not split, so that the global dimension is in fact equal to 1.
Question 2 seems more difficult to answer, but I would certainly be interested to know any information, even simply partial answers! It seems to me that the sequence for $M$ above is still exact for general $R$, $V$, and $M$. (I could be slightly off, as the Cuntz-Quillen results assume that all rings are algebras over a field, but I suspect their proofs are likely to still work.) In this case, I wonder whether in certain situations we may be able to control how the global dimensions behave when passing from $M$ (considered as a left $R$-module) to the $A$-modules $A \otimes_R M$ and $A \otimes_R V \otimes_R M$.
