In continuous time, in general you need the jump matrix in addition to the generator matrix. The jump matrix will tell you the probability of each path from $i$ to $j$, while the generator will then tell you the distribution of the holding times for each step of a specified path.
Assuming none of rows of the generator are zero, the jump matrix $\Pi$ has off-diagonal entries $\Pi_{ij}=\frac{L_{ij}}{\sum_{j=1,j \neq i}^n L_{ij}}$, and diagonal entries of zero. So your generator has a jump matrix of
$$\Pi = \begin{bmatrix} 0 & 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 0 & 1 & 0 \end{bmatrix}.$$
In the particular case you asked, one always transitions from $3$ to $2$, so the hitting time for $2$ from $3$ is just the holding time at $3$. In a continuous time Markov chain the holding time at $i$ is exponentially distributed with parameter $-\frac{1}{L_{ii}}$. Intuitively, $-L_{ii}$ measures the frequency of transitions, so its reciprocal measures the time of transitions. There is a natural analogy to electric networks here: at a node $i$ you have $n-1$ parallel resistors extending to each state $j$, each of which has resistance $\frac{1}{L_{ij}}$ (which is understood as $+\infty$ if $L_{ij}=0$).
It is an interesting exercise to prove that the holding times are exponentially distributed. It follows from the more elementary fact that the only random variables with the "memoryless property" $P(X>t+s | X>s)=P(X>t)$ are exponential variables.
For the second question, there are a couple possible paths, which you might describe as a tree. Let's instead assume that you start at 1 with probability 1. Now you will either go to 2 with probability 1/2 or go to 3 with probability 1/2. If you do the latter then you will go to 2 with probability 1 after that. So there are two possible paths, each with probability 1/2. If you take the first one, you arrive in a time which is exponentially distributed with parameter $\frac{1}{2}$ (corresponding to the holding time at 1). If you take the second one, you arrive in a time which is exponentially distributed with parameter $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ (corresponding to the holding time at 1 followed by the holding time at 3). Do you see how to combine these?
