My question is whether it is possible to compute hitting times from the previously calculated stationary distribution $\pi$ of a continuous-time Markov process $(X_t)_{t \geq 0}$. I know that, from learning about Markov chains in discrete time, $$\pi(j) = \frac{1}{\mathbb{E}_j(T_j)} \,\,\, \forall j\,$$ (such that $T_j$ is the return time to state $j$). But, is there any way to compute $\mathbb{E}_i(T_j)$ for $i \neq j$ from $\pi$ in the discrete time case? Or, more generally, in the continuous time case, if we have some subset $A \subset \mathbb{S}$ of a (finite) state space, then is there a way to compute $\mathbb{E}_i(T_A)$ directly from (or at least using) the stationary distribution $\pi$, where $T_A = \inf\{t:X_t \in A\}$ and $i \notin A$ ?
Let me give my specific example to focus on in order to further clarify. Consider a Markov process with state space (ordered) $\mathbb{S}=\{0,1,2,12\}$ and a generator matrix (with rows and columns in the same order)
$$Q =\begin{pmatrix} -2 & 2 & 0 & 0 \\ 0 & -4 & 4 & 0 \\ 2 & 0 & -4 & 2 \\ 0 & 2 & 4 & -6 \end{pmatrix} \,.$$
Computing the stationary distribution, we obtain $\pi = (\pi_0, \pi_1, \pi_2, \pi_{12}) = (\frac{1}{3},\frac{2}{9},\frac{1}{3},\frac{1}{9})$. Now, we can set $g(i) = \mathbb{E}_i(T_0)$ for $i \neq 0$, $g(0) = 0$, and solve the following system of three equations $(i \neq 0 \implies i = 1,2,12)$ for $g(i)$: $$\sum_{j \in \mathbb{S}} Q(i,j)g(j) = -1\,.$$
Doing this, we obtain $g = (g(0),g(1),g(2),g(12)) = (0,1,\frac{3}{4},1)$. My question is if, for instance, we could compute $g(12)$ directly from the stationary distribution $\pi$.