= Solution
On the negative branch set $Y=-X$. Its outward step probability is $3/4$ and inward probability is $1/4$. The same finite <gambler's ruin> recurrence as in part (a), with root $(1/4)/(3/4)=1/3$, gives $\Pr_{-1}(T_0<\infty)=1/3$. Hence, following one departure from zero, the probability of returning is
$$
r=\frac12\cdot\frac12+\frac12\cdot\frac13=\frac5{12}<1.
$$
At each return the <Strong Markov property> restarts the same excursion experiment. The probability of at least $m$ returns is $r^m$, tending to zero. Thus almost surely there is a last visit to zero.
After that visit the walk remains strictly positive or strictly negative, because nearest-neighbour steps cannot change sign without crossing zero. On the positive branch it can be coupled, up to any return, to independent steps of mean $2/3-1/3=1/3$. The <strong law of large numbers> gives a walk tending to $+\infty$ almost surely. This conclusion remains true on the event that the excursion never returns: conditioning on a positive-probability event cannot make a probability-zero failure of the law occur. On the negative branch the mean step is $1/4-3/4=-1/2$, giving $-\infty$ in the same way. Apply this at each possible departure time, a countable family, or use the <Strong Markov property> at the successive returns. Therefore
$$
\boxed{\Pr_0(X_n\to+\infty\ \hbox{or}\ X_n\to-\infty)=1.}
$$
Finite visits to zero alone would not prove divergence on an arbitrary chain; the outward-drift argument is the additional step needed here.
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