No to both questions. Consider $V=\{ g\in\mathcal S: g(-x)=ig(x) \}$. This space is closed because it is the null space of the continuous operator $R-i$, with $(Rg)(x)=g(-x)$.

Then $T_f=0\in V'$ for every $f\in V$ because $fg$ is odd when $f,g\in V$.

No to both questions. Consider $V=\{ g\in\mathcal S: g(-x)=ig(x) \textrm{ for }x\ge 0\}$. This space is closed and $V\not= 0$ because we can start with a compactly supported function on $[0,\infty)$, with support at some distance from $x=0$ and then define $g$ on $(-\infty, 0)$ as required.

Then $T_f=0\in V'$ for every $f\in V$ because $fg$ is odd when $f,g\in V$.