Let $a_{i}$ denote the abosorption probabilites into state $4$ starting from $i$ \begin{align} a_{5} &= 0, a{4} = 1 \newline a_{i} &= \sum_{j} a_{j}p_{ij}\newline a_{2} &= a_{1}p_{21} + a_{4}p_{24}\newline a_{3} &= a_{1}p_{31} + a_{2}p_{32} + a_{5}p_{35}\newline a_{1} &= a_{2}p_{12} + a_{3}p_{13} \end{align} Solving, $a_{1} = \frac{9}{14}, a_{2} = \frac{5}{7}$ and $a_{3} = \frac{15}{28}$

Let $\mu_{i}$ denote the expected time till absorption starting from $i$, then \begin{align} \mu_{4} &= 0 \newline \mu_{1} &= 1 + \mu_{2}p_{12} + \mu_{3}p_{13} \newline \mu_{2} &= 1 + \mu_{1}p_{21} + \mu_{4}p_{24} \newline \mu_{3} &= 1 + \mu_{1}p_{31} + \mu_{2}p_{32} \end{align} Solving, $\mu_{1} = \frac{55}{4}, \mu_{2} = 12$ and $\mu_{3} = \frac{111}{8}$