= Solution
All four permitted hops have rate one: $1\to2$, $2\to1$, $2\to3$, and $3\to1$. Every state reaches every other, so \b[the only <communicating class> is $\{1,2,3\}$], and it is closed.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-1-frog-chain.png]
{title=Allowed frog hops and their rates}
Solving $\pi Q=0$, $\sum_i\pi_i=1$, gives $\pi_1=2\pi_2$ and $\pi_3=\pi_2$. Thus
$$
\boxed{\pi=(1/2,1/4,1/4),\qquad\mathbb P_\pi(X=2)=1/4.}
$$
For the transient <probability>, let $p_j(t)=P_{1j}(t)$. The forward equations and $p_1+p_2+p_3=1$ give $p_2'=p_1-2p_2$, $p_1'=1-2p_1$. From $p_1(0)=1$, $p_1(t)=1/2+e^{-2t}/2$. Integrating $p_2'+2p_2=p_1$, with $p_2(0)=0$, yields
$$
\boxed{P_{12}(t)=\frac14\left[1+(2t-1)e^{-2t}\right].}
$$
It has <derivative> one at zero and tends to $1/4$, as required.
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