I give it a try with your first question: If $A\in\mathbb{R}^{m\times n}$ then $$\frac{1}{\sqrt{n}}\|A\|_\infty\leq \|A\|_2\leq \sqrt{m}\|A\|_\infty$$ $$\frac{1}{\sqrt{m}}\|A\|_1\leq \|A\|_2\leq \sqrt{n}\|A\|_1$$
Also $$\|AB\|_a\leq \|A\|_a\|B\|_a$$
Note that $\|B_i\|_2<1$ implies that absolute value of the every element of matrix B is less than 1, $|B_{ij}|<1$, then $\|B\|_\infty (infinity norm of a matrix is the maximum absolute row sum of the matrix). And with the same argument $\|B\|_1.
Then using these properties we have $$\|vA^TB\|_\infty\leq \sqrt{n}\|vA^TB\|_2\leq \sqrt{n} \|vA^T\|_2\|B\|_2\leq \sqrt{nm} \|vA^T\|_2\|B\|_\infty\leq n\sqrt{nm} \|vA^T\|_2$$ or $$\|vA^TB\|_\infty\leq \sqrt{n}\|vA^TB\|_2\leq \sqrt{n} \|vA^T\|_2\|B\|_2\leq n \|vA^T\|_2\|B\|_1\leq nm \|vA^T\|_2$$ (these are conservative bounds you can probably obtain a tighter bound)