= Solution
After the first step to $+1$, the right-hand walk moves away with probability $3/4$ and toward zero with probability $1/4$. Part (ii) therefore gives return probability $1/3$ from that side. After a first step to $-1$, reverse the sign of the coordinate: movement away from zero has probability $2/3$ and toward it probability $1/3$, giving return probability $1/2$. Weight by the probabilities of the first step from zero:
$$
\boxed{P_0(T_0^+<\infty)=\frac13\frac13+\frac23\frac12=\frac49.}
$$
There is no holding step at zero, so every positive-time return is included by this decomposition.
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