Secure blinding factor switching at malicious server-side (Switching in One Time pad)
I'll answer question 2, leaving the first as an exercise to the reader. I'll do this on intuitive grounds, rather than using explicit conditional probabilities.

The adversary is free to compute $v_1\cdot v_2$ regardless of what we ask, therefore removing everything about that and $v_3$ does not change the problem, which reduces to:


  We somewhat have chosen some $a\in\mathbb F_p$. We draw a random uniform secret $z_1\in\mathbb F_p^*$ (so that it's inverse ${z_1}^{-1}$ is well-defined) and a random uniform secret $z_2\in\mathbb F_p$, and compute and reveal $v_1=a\cdot z_1$ and $v_2={z_1}^{-1}\cdot z_2$ to the adversary; what does that reveal about $a$?


Lemma 1: for unknown $u\in\mathbb F_p^*$, drawing a random uniform secret $z\in\mathbb F_p$, and revealing $v=u\cdot z$, reveals nothing about $u$.

Lemma 2: for unknown $u\in\mathbb F_p^*$, drawing a random uniform secret $z\in\mathbb F_p^*$, and revealing $v=u\cdot z$, reveals nothing about $u$.

Proof follows from the fact that $z\to u\cdot z$ is a mapping over $\mathbb F_p$ (for lemma 1) or over $\mathbb F_p^*$ (for lemma 2).

Notice that neither $v_2$ nor $z_2$ are involved when we compute and reveal $v_1=a\cdot z_1$. Therefore, we can consider in isolation the part of the protocol where we draw a random uniform secret $z_2\in\mathbb F_p$ and reveal $v_2={z_1}^{-1}\cdot z_2$. We apply lemma 1 with $u={z_1}^{-1}$ (which belongs to $\mathbb F_p^*$), and conclude that revealing $v_2$ reveals nothing about ${z_1}^{-1}$, hence nothing about $z_1$.

Thus the part of the protocol where we compute and reveal $v_2={z_1}^{-1}\cdot z_2$ has revealed nothing about any quantity in the part of the protocol where we compute and reveal $v_1=a\cdot z_1$. If our choice of $a$ was not zero, by lemma 2, that part of the protocol has revealed nothing about $a$. If our choice of $a$ was $0$, $v_1$ will be $0$.

Hence the answer to question 2 is: the protocol reveals precisely whether $a=0$ or not. No other information about $a$ leaks.

With the statement disallowing $z_1=0$ (or if we reject that as having vanishing odds since $p$ is large), it can be shown that no (or vanishingly few) information about $z_1$ in isolation leaks, and that the only (or almost the only) information that leaks about $z_2$ in isolation is whether $z_2=0$ hold.