In general topology, it is often the case that theorems that rely on Hausdorffness can be sometimes generalized to the non-Hausdorff setting by imposing regularity as a separation axiom. Here is a theorem that solves Question 2. Note also that this solves Question 1, since then necessarily $Y$ must be completely regular (pseudometrizable spaces are completely regular).
Theorem: If $X$ is compact pseudometrizable, $Y$ is regular and $f : X \to Y$ is continuous and surjective then $Y$ is pseudometrizable.
I follow the proof in [Aliprantis, Border, "Infinite Dimensional Analysis: A Hitchhiker's Guide", Corollary 3.43] for the statement "$X$ compact metrizable, $Y$ Hausdorff, $f$ continuous surjective $\Rightarrow$ $Y$ metrizable" with the obvious modifications replacing non-closed compact sets with their closures.
Proof: Since $X$ is compact and $f$ is a continuous it follows that $Y = f(X)$ is compact. Since $Y$ is compact and regular it follows that $Y$ is also completely regular and normal. By the Urysohn metrization theorem in the non-Hausdorff setting, every regular second-countable topological space is pseudometrizable. Thus it is enough to show that $Y$ is second-countable. If $G \subseteq X$ is open then $X \setminus G$ is closed and thus compact (as a closed subset of a compact space). Therefore, $f(X \setminus G)$ is compact, but it need not be closed in $Y$. However, its closure $\overline{f(X \setminus G)}$ in $Y$ is compact since $Y$ is regular (see here). Thus, if $G$ is open in $X$ then $Y \setminus \overline{f(X \setminus G)}$ is open in $Y$. Since $X$ is compact and pseudometrizable it follows that $X$ is second-countable. So let $\mathcal{B}$ be a countable base for $X$ and we can assume that $\mathcal{B}$ is closed under finite unions. Define $\mathcal{C} := \{ Y \setminus \overline{f(X \setminus G)} \mid G \in \mathcal{B} \}$. Then $\mathcal{C}$ is countable an we claim that $\mathcal{C}$ is a base for $Y$. Let $W \subseteq Y$ open, $W \neq \emptyset$. Take $y \in W$. Then $\{ y \}$ is compact and again by regularity of $Y$, $W$ is also an open neighborhood of its closure $\overline{ \{ y \} }$ (see again here). Then $f^{-1}(\overline{ \{ y \} })$ is closed in $X$, thus compact and we have $f^{-1}(\overline{ \{ y \} }) \subseteq f^{-1}(W)$. Now since $X$ is second-countable the open set $f^{-1}(W)$ is a union of sets in $\mathcal{B}$ which then forms an open cover of $f^{-1}(\overline{ \{ y \} })$ and thus there are finitely many such sets $B_1, \dots, B_n \in \mathcal{B}$ with $f^{-1}(\overline{ \{ y \} }) \subseteq G \subseteq f^{-1}(W)$ where $G := \bigcup_{i=1}^n B_i$. Since $\mathcal{B}$ is closed under finite unions we have $G \in \mathcal{B}$. It follows that $Y \setminus \overline{f(X \setminus G)} \subseteq f(G) \subseteq W$ and we are done.
