= Solution
The chain is a <reversible Markov chain>, so detailed balance makes the stationary path probability
$$
\pi_{i_0}P_{i_0i_1}\cdots P_{i_{m-1}i_m}
=\pi_{i_m}P_{i_mi_{m-1}}\cdots P_{i_1i_0}.
$$
A path contains an ordered occurrence of $a$ followed later by $b$ exactly when its reversal contains $b$ followed later by $a$. These occurrences need not be adjacent. Therefore $\Pr(\tau(a,b)\le m)=\Pr(\tau(b,a)\le m)$ for every $m$, and in particular the two finite expectations are equal. This is <ordered-state visit times in a reversible chain>.
To decompose the expectations, distinguish the entrance time $h(i,j)$, which allows time zero and has $h(j,j)=0$, from the positive return convention $k(j,j)$ in part (i). The <Strong Markov property> at the first visit to $a$ or $b$ gives
$$
E\tau(a,b)=\sum_i\pi_i h(i,a)+k(a,b),\qquad
E\tau(b,a)=\sum_i\pi_i h(i,b)+k(b,a).
$$
Equating them yields the difference formula with $h$. Replacing $h$ by $k$ in the sum adds $\pi_a k(a,a)-\pi_b k(b,b)=1-1=0$, again by <Kac's lemma>. Hence
$$
\boxed{k(b,a)-k(a,b)=\sum_i\pi_i[k(i,a)-k(i,b)].}
$$
For a numerical consistency check, the other entrance times are $k(b,a)=15/11$ and $k(c,a)=16/11$, while $k(b,b)=3$. Both sides of the displayed identity equal $-6/11$.
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