I have just realized that the first question has a simple affirmative answer.
Theorem 1. Every countable first-countable $T_1$ space $X$ is symmetrizable.
Proof. For every $x\in X$, fix a neighborhood base $(U_n(x))_{n\in\omega}$ such that $U_{n+1}(x)\subseteq U_n(x)$ for all $n\in\omega$. Since $X$ is countable, there exists a linear order $\le$ on $X$ such that for every $x\in X$ the initial interval $\{y\in X:y\le x\}$ is finite.
It is easy to see that the topology of $X$ is generated by the symmetric
$$d(x,y)=\inf\big\{2^{-n}:\max\{x,y\}\in U_n(\min\{x,y\})\big\}.\quad\square$$
Important Remark. After looking at a reference Dave L. Renfro suggested in a comment as possibly being relevant, I discovered that affirmative answers to both my questions follow from
Theorem 2.9 (Arhangelski, 1966): Every first-countable $T_1$ space with a $\sigma$-discrete (closed) network is symmetrizable.
But reading the proof of this theorem I discovered that it works only for regular spaces. Arhangelski writes that one loses no generality assuming that the $\sigma$-discrete network is closed, i.e., consists of closed sets. But after taking the closure of elements of a network in a $T_1$-space (even in a Hausdorff space), the network property can be destroyed. So, Question 2 seems to stay open and not answered even for Hausdorff (not mention $T_1$) spaces.
Added in Edit 31.05.2022. To my big surprise (and contrary to what was claimed by Arhangelski in his Theorem 2.9), I have just discovered that Question 1 has negative answer!
First let us prove that the symmetrizability of second-countable spaces is equivalent to the perfectness. Recall that a topological space $X$ is perfect if every closed subset of $X$ is of type $G_\delta$.
Theorem 2. A second-countable (Hausdorff) $T_1$-space $X$ is symmetrizable if (and only if) it is perfect.
Proof. If $X$ is symmetrizable and Hausdorff, then the first-countability of $X$ implies that $X$ is semi-metrizable and perfect, see the paragraph before Theorem 9.8 in Gruenhage's "Generalized metric spaces".
Conversely, if a second-countable $T_1$-space $X$ is perfect, then each open subset of $X$ is an $F_\sigma$ in $X$, which implies that $X$ has a countable closed network. Now we can apply Arhangelski's Theorem 2.9 to conclude that $X$ is symmetrizable. $\square$
Let $\mathrm{non}(\mathcal M)$ denote the smallest cardinality of a nonmeager set in the real line.
Example 1. There exists a second-countable Hausdorff space of cardinality $\mathrm{non}(\mathcal M)$ which is not perfect and hence not symmetrizable.
Proof. Take any nonmeager linear subspace $L$ of cardinality $\mathrm{non}(\mathcal M)$ in $\mathbb R^\omega$ such that for every $n\in\omega$ the intersection $L_n=L\cap(\{0\}^n\times\mathbb R^{\omega\setminus n})$ is dense in $\{0\}^n\times\mathbb R^{\omega\setminus n}$. Consider the quotient space $X=L_\circ/_\sim$ of $L_\circ=L\setminus\{0\}$ by the equivalence relation $\sim$ defined by $x\sim y$ iff $\mathbb R x=\mathbb Ry$. Since the space $L_\circ$ is Baire and the quotient map $q:L_\circ\to X$ is open, the space $X$ is second-countable and Baire. It is easy to check that the closure of every nonempty set in $X$ contains the set $q[L_n\setminus\{0\}]$ for some $n\in\omega$. This implies that the space $X$ is superconnected in the sense that for every nonempty open sets $U_1,\dots,U_n$ in $X$ the intersection of their closures $\overline U_1\cap\dots\cap\overline U_n$ is not empty.
Now take any disjoint nonempty open sets $U,V$ in $X$. Assuming that $V$ is of type $F_\sigma$, we can apply the Baire Theorem and find a nonempty open set $W\subseteq V$ whose closure in $X$ is contained in $V$. Then $\overline{U}\cap\overline{W}=\emptyset$, which contradicts the superconnectedness of $X$. $\square$
The cardinality $\mathrm{non}(\mathcal M)$ is the above example can be lowered to $\mathfrak q_0$, where $\mathfrak q_0$ is the smallest cardinality of a second-countable metrizable space which is not a $Q$-space (= contains a subset which is not of type $G_\delta$).
A topological space is submetrizable it it admits a continuous metric.
Each submetrizable space is functionally Hausdorff in the sense that for any distinct elements $x,y\in X$ there exists a continuous function $f:X\to\mathbb R$ such that $f(x)\ne f(y)$.
Example 2. There exists a submetrizable second-countable space $X$ of cardinality $\mathfrak q_0$, which is not symmetrizable.
Proof. By the definition of the cardinal $\mathfrak q_0$, there exists a second-countable metrizable space $Y$, which is not a $Q$-space and hence contains a subset $A$ which is not of type $G_\delta$ in $X$. Let $\tau'$ be the topology on $X$, generated by the subbase $\tau\cup\{X\setminus A\}$ where $\tau$ is the topology of the metrizable space $Y$. It is clear that $X=(Y,\tau')$ is a second-countable space containing $A$ as a closed subset. Since $\tau\subseteq\tau'$, the space $X$ is submetrizable. Assuming that $X$ is symmetrizable and applying Theorem 2, we conclude that $X$ is perfect and hence the closed set $A$ is equal to the intersection $\bigcap_{n\in\omega}W_n$ of some open sets $W_n\in\tau'$. By the choice of the topology $\tau'$, for every $n\in\omega$ there exists open sets $U_n,V_n\in \tau$ such that $W_n=U_n\cup(V_n\setminus A)$. It follows from $A\subseteq W_n=U_n\cup(V_n\setminus A)$ that $A=A\cap W_n=A\cap U_n\subseteq U_n$.
$$A=\bigcap_{n\in\omega}W_n=A\cap\bigcap_{n\in\omega}W_n=\bigcap_{n\in\omega}(A\cap W_n)=\bigcap_{n\in\omega}(A\cap U_n)\subseteq \bigcap_{n\in\omega}U_n\subseteq \bigcap_{n\in\omega}W_n=A$$
and hence $A=\bigcap_{n\in\omega}U_n$ is a $G_\delta$-set in $X$, which contradicts the choice of $A$. This contradiction shows that the submetrizable second-countable space $X$ is not symmetrizable. $\square$
On the other hand we have the following partial affirmative answer to Question 1.
Theorem 3. Martin's Axiom implies that every second-countable $T_1$ space of cardinality $<\mathfrak c$ is perfect and hence symmetrizable.
Proof. It is known that Martin's Axiom implies that every second-countable $T_1$-space $X$ of cardinality $\mathfrak c$ is a $Q$-space, which means that every subset of $X$ is of type $G_\delta$. In particular, $X$ is perfect and by Theorem 2 is symmetrizable. $\square$
Best Answer
Every second-countable $T_0$ space is a quotient of a separable metric space and this essentially follows from the proof of the $T_1$ case given by Paul Strong in Quotient and pseudo-open images of separable metric spaces [Proc. Amer. Math. Soc. 33 (1972), 582-586].
A continuous map $f:X \to Y$ is sequence covering if for every convergent sequence $y_n \to y$ in $Y$ there is a convergent sequence $x_n \to x$ in $X$ such that $y_n = f(x_n)$ for all $n$ and $y = f(x)$. (Since the spaces under consideration are not necessarily Hausdorff, when I say "convergent sequence" I always mean a convergent sequence together with a choice of limit.)
Fact 1. If $f:X \to Y$ is a sequence covering continuous map and $Y$ is sequential then $f:X \to Y$ is a quotient map.
To prove this, we need to show that a map $g:Y \to Z$ is continuous if (and only if) $g \circ f:X \to Z$ is continuous. Since $Y$ is sequential, it is enough to show that if $y_n \to y$ is a convergent sequence in $Y$, then $g(y_n) \to g(y)$ is a convergent sequence in $Z$. Since $f:X \to Y$ is sequence covering, we can find a convergent sequence $x_n \to x$ in $X$ that maps to $y_n \to y$. Assuming only that $g \circ f:X \to Z$ is continuous, it follows that $g(f(x_n)) = g(y_n) \to g(f(x)) = g(y)$ is indeed a convergent sequence in $Z$.
Let $\omega+1 = \{0,1,\dots,\omega\}$ be the one-point compactification of $\omega$. The space $Y^{\omega+1}$ with the compact-open topology consists of all convergent sequences in $Y$ (along with a choice of limit). Since $\omega+1$ is compact Hausdorff, the evaluation map $e:(\omega+1)\times Y^{\omega+1}\to Y$ is continuous. Moreover, this map is obviously sequence covering. Even better:
Fact 2. If $g:X \to Y^{\omega+1}$ is a continuous surjection then the continuous map $f:(\omega+1)\times X \to Y$ defined by $f(n,x) = e(n,g(x)) = g(x)(n)$ is sequence covering.
Combining these facts, we see that if $Y^{\omega+1}$ is sequential and the continuous image of a separable metric space $X$ then $Y$ is a quotient of the separable metric space $(\omega+1)\times X$.
Now if $Y$ is $T_0$ and second-countable then so is the space $Y^{\omega+1}$. Therefore, it suffices to prove:
Fact 3. Every second-countable $T_0$ space $Y$ is the continuous image of a separable metric space.
Tho see this, fix a countable base $(U_n)_{n\lt\omega}$ for $Y$. Let $X \subseteq \omega^\omega$ consist of all $x:\omega\to\omega$ such that $(U_{x(n)})_{n \lt\omega}$ enumerates a neighborhood basis for some point of $Y$. Since $Y$ is $T_0$, a neighborhood basis for a point uniquely determines that point. Therefore, we obtain a natural surjection $f:X \to Y$, which is easily seen to be continuous.
Since a second-countable space is always sequential, we can combine the three facts above to conclude that every second-countable $T_0$ space is a quotient of a separable metric space.
Actually, as pointed out in Andrej's answer the map in Fact 3 is already a quotient map (since it is an open mapping). Strong needed the more complicated construction since his assumptions are in terms of countable networks instead of countable bases.