It can happen that $Pic(X) = Pic(Y) = Pic(Z) = 0$ but $Pic(W) \neq 0$!

For example, let $f: \mathcal{E} \to Z$ be a non-isotrivial family of elliptic curves, where $Z$ is a smooth rational curve. Then $Pic(\mathcal{E})$ is finitely generated, so by removing a finite number of (images of) sections of $f$ we obtain a surface $X \to Z$ such that $Pic(X) = 0$.
Letting $Y = X$, I claim that $Pic(W) \neq 0$:

To see this, let $\Delta$ be the diagonal inside $W = X \times_Z X$ so $L :=\mathcal{O}(\Delta) \in Pic(W)$. The restriction of $L$ to any fibre of the map $W \to Z$ is nonzero
since the cohomology class of the diagonal in $E \times E$, where $E$ is any elliptic curve, remains non-zero when restricted to $E' \times E'$ where $E' \subset E$ is any non-empty Zariski open subset. In particular, $L \neq 0$, so $Pic(W) \neq 0$.

3) is also false.

For example, let $Y \to Z$ be a finite etale cover such that $Z$ is a smooth rational curve and the genus of $\bar{Y}$, the smooth compactification of $Y$, is at least $2$. Let $X$ also be a rational curve and take any morphism $X \to Z$ of degree $> 1$. Then $W$ is a smooth curve with a map to $Y$ of degree $> 1$. Since $g(\bar{Y}) > 1$, at least one component of $\bar{W}$ has genus $> g(\bar{Y})$ or it has more than one component, so the cokernel of the induced map $Pic(\bar{Y}) \to Pic(\bar{W})$ is uncountable. Since the kernel of the map $Pic(\bar{W}) \to Pic(W)$ is finitely generated, it follows that the the map $Pic(Y) \to Pic(W)$ cannot be surjective.