Electrochemistry - reduction potentials - an IB Chemistry Question
First put your redox potentials on a horizontal line by order of increasing standard redox potentials. You will see that it starts on the left-hand side by $\ce{Zn^{2+}/Zn}$ at $\pu{-0.76 V}$. Then, going to the right, you find $\ce{V^{III}/V^{II}}$ at $\pu{-0.26 V}$. Now you find $\ce{Pb^{2+}/Pb}$, and so on.
Now try to learn how to use this scale. The reduced species of any couple can react with any oxidized species of any couple situated to its right. For example, $\ce{Zn}$ is the reduced species of the couple $\ce{Zn^{2+}/Zn}$. So $\ce{Zn}$  can reduce all oxidized species, and for example all oxidized forms of Vanadium ions. But $\ce{Zn^{2+}}$ cannot be reduced, because there is no reduced species at the left hand side of $\pu{- 0.76 V}$.
On the other end of the scale, you find $\ce{Cl2/Cl-}$ at $\pu{+1.36 V}$. Hopefully you admit that $\ce{Cl2}$ can oxidize all reduced species, and $\ce{Cl-}$ cannot reduce anything.
Now let's consider $\ce{Pb^{2+}/Pb}$, at $\pu{-0.13 V}$. $\ce{Pb}$ can reduce all oxidized species situated on the positive side of the axis, for example $\ce{VO^{2+}}$ which is at $\pu{+0.34 V}$. But it cannot reduce $\ce{V^{3+}}$ to $\ce{V^{2+}}$, because $\ce{V^{3+}/V^{2+}}$ is on tis left-hand side. And this is exactly your first question.
Now think a little bit about how to solve the second question. It is exactly the same approach.