I have the following assignment:
For part a. I found that there are 3 communicating classes: $\{0,1\}$ (closed, recurrent), $\{2\}$ (closed, recurrent) and $\{3\}$ (not closed, transient). For part b. I found that $X_H$ is uniformly distributed on $\{0,1,2\}$ with probability $\frac{1}{3}$. It is part c. that I do not know how to solve. Since the chain is not irreducible, it does not converge to a unique invariant distribution, so I cannot solve $\pi P = \pi$. I know I have to do something with the result of part b, since it's again conditioned on $X_0 = 3$, but I don't know what.
The idea is that you start in 3, but at some point move to one of the closed communicating classes. So you can condition on $X_H$, which determines in which communicating class you end up. In case $i=0$ we e.g. have: \begin{align*} \lim_{n \to \infty} \mathbb{P}(X_n = 0 | X_0 = 3) = \lim_{n \to \infty} \sum_{j = 0}^2 \frac{1}{3}\mathbb{P}(X_n = 0 | X_0 = 3, X_H = j) = \frac{1}{3}(\mathbb{P}(X_n = 0 | X_H = 0) + \mathbb{P}(X_n = 0 | X_H = 1)). \end{align*} These last two probabilities can be computed using $\pi P' = \pi$ with $P'$ the 2x2 upper left part of $P$. The same technique can be used in case i = 1,2. Especially in case $i = 2$ the results easily follow, since that class has one single item: \begin{align*} \lim_{n \to \infty} \mathbb{P}(X_n = 2 | X_0 = 3) = \lim_{n \to \infty} \sum_{j = 0}^2 \frac{1}{3}\mathbb{P}(X_n = 2 | X_0 = 3, X_H = j) = \frac{1}{3}\mathbb{P}(X_n = 2 | X_H = 2) = \frac{1}{3}. \end{align*}