How to calculate return time from a transition probability matrix? 
Consider the transition probability matrix with state space $\{0,1,2,...5\}$ as follows
$$P= \begin{bmatrix} 1/2 & 1/2 & 0 & 0 & 0 & 0 &\\
1/3 & 2/3   & 0 & 0 & 0 & 0\\
0 & 0 & 1/8 & 0  & 7/8 & 0\\
1/4 & 1/4 & 0 & 0 & 1/4 & 1/4\\
0 & 0 & 3/4 & 0 & 1/4 & 0\\
0 & 1/5 & 0 & 1/5 & 2/5 & 1/5\end{bmatrix}$$
a) Let $C=\{0,1\}$. Then find $\rho_{3}(C)$ i.e., absorption probability to the set $C$ starting from $3$.
b) Find $m_0=\mathbb{E}_0(T_0)$, where $\mathbb{E}_i(T_i) := \mathbb{E}(T_i \mid X_0=i)$

Here is what I have thought of.
I know that $\{0,1\}$ is a recurrent irreducible closed set and $\{3,5\}$ is another irreducible closed set. Hence starting from $3$, it would be impossible to go to any other state except $5$ and so the absorption probability of  $\rho_3(C)$ must be $0$. Coming to the second part, I am not sure how to exactly do it? We haven't explicitly learnt how to calculate hitting time, and so I don't know how to start calculating the conditional expectation using this. I also read that $\mathbb{E}_i [T_i]=m_i=\frac{1}{\pi_i}$ where $\pi_i$ is the stationary distribution. But $P$ isn't an irreducible Markov Chain so it doesnt have a unique stationary distribution. So I don't know how to proceed using that formula either. I am unaware of any other formula to calculate return time, so any help would be appreciated!
 A: Your understanding of absorption probailities is slightly incorrect. Here is a formula for absorprtion probabilities-
$$\rho_C(x)= \sum_{y \in C} P(x,y) + \sum_{y \in \mathcal{S}_{T}} P(x,y) \rho_C(y)$$
Given that $C=\{0,1\}$ and $\mathcal{S}_T = \{3,5\}$, we have
\begin{align}
    \rho_C(3) &=P(3,0)+P(3,1)+P(3,3)\rho_C(3) + P(3,5) \rho_C(5)\nonumber\\
    &= 1/4 + 1/4 + 1/4 \rho_C(5)\nonumber\\
    \implies &\rho_c(3)- 1/4 \rho_C(5) = 1/2
\end{align}
and
\begin{align}
    \rho_C(5) &=P(5,0)+P(5,1)+P(5,3)\rho_C(3) + P(5,5) \rho_C(5)\nonumber\\
    &= 1/5 + 1/5 \rho_C(3) + 1/5 \rho_C(5)\nonumber\\
    \implies & 4/5\rho_C(5)- 1/5 \rho_C(3) = 1/5 
\end{align}
Then, solving the two equations, we get $\rho_C(3)=3/5$ and $\rho_C(5)=2/5$.
For $m_0$, consider the irreducible closed classes in this M.C. We have $\pi P= \pi \implies P^T \pi^T= \pi^T$. Thus,
$$ \left[\begin{array}{cc}
1/2 & 1/3 \\
1/2 & 2/3
\end{array}\right]
\left[\begin{array}{c}
\pi_0 \\
\pi_1 \\
\end{array}\right] 
=
\left[\begin{array}{c}
\pi_0 \\
\pi_1 \\
\end{array}\right]
$$
Now we have the equations
$$1/2 \pi_0 + 1/3\pi_1= \pi_0 \Rightarrow -1/2 \pi_0+ 1/3 \pi_1=0$$
$$ \pi_0 +\pi_1=0$$
Solving this yields $\pi(0) = 2/5$ and $\pi(1) = 3/5$.
