= Solution
Let $w$ count the unequal neighbouring pairs, or <domain walls>. The <energy> is $E=-JL+Jw$. A single <domain wall> is impossible with <periodic boundary conditions>: following all the other, equal-state bonds around the ring would force its endpoints to be equal too. Two <domain walls> are possible, splitting the ring into two nonempty uniform arcs carrying different states. Thus the <first excited state> has
$$
\boxed{E_1=-JL+2J,\qquad E_1-E_0=2J.}
$$
Choose the two wall bonds in $\binom L2$ ways. For that choice, identify one of the resulting arcs by its smallest site label. Its state has $Q$ choices and the other arc has $Q-1$. This counts each configuration exactly once, giving the <two-domain-wall degeneracy of a Potts ring>
$$
\boxed{d_1=\binom L2Q(Q-1).}
$$
For example, $Q=2,L=4$ gives $d_0=2$ and $d_1=12$; the remaining two configurations are alternating and have four <domain walls>. The formula also covers $L=2$, where the specified ring <energy> counts the two directed neighbouring bonds, and an unequal pair breaks both. If $Q=1$ or $L=1$, every allowed configuration is a <ground state> and there is no <first excited state>. For $Q\ge3$, the number of <domain walls> need not be even: three distinct states on a three-site ring give three, although one remains impossible.
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