= Solution
Within the valid parameter rectangle, suppose $a>0$ and $b<1/3$. The chain is then irreducible by part (i). State $1$ has a two-step return $1\to2\to1$ and a three-step return $1\to2\to4\to1$, both of positive <probability>. Its <period of a state in a Markov chain> therefore divides both two and three, hence equals one. Irreducibility gives the same period to every state, so the chain is aperiodic.
If $b=1/3$ and $a>0$, every transition goes between the two classes $\{1,3\}$ and $\{2,4\}$. Returns must have even length, while $1\to2\to1$ has length two. Thus the irreducible chain has period two.
If $a=0$, the closed communicating class $\{1,3,4\}$ alternates between $\{1,3\}$ and $\{4\}$. Its states have period two, using the positive two-step returns through $4$. Consequently the chain is not aperiodic, regardless of how the transient state $2$ is classified. Combining the cases,
$$
\boxed{\text{Aperiodic exactly when }0<a\leq1,\quad0\leq b<\frac13.}
$$
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