= Solution
Fix $M$. Each time the chain is in $\{1,\ldots,M\}$ immediately before a jump, its conditional chance to jump to zero is at least the $\delta$ from part (b). Count all these opportunities, including returns to the same state after an excursion; the $T_m$ in part (b) need not count those returns. By the <Strong Markov property> at successive such visits, the probability of surviving $r$ opportunities without absorption is at most $(1-\delta)^r$. Hence infinitely many such opportunities without hitting zero have probability zero.
A positive state cannot be occupied forever: its finite positive holding rate gives an almost surely finite holding time. For a <nonexplosive continuous-time Markov chain>, visits at arbitrarily late times to the finite set therefore require infinitely many of the counted opportunities. On the event of no absorption, the chain consequently eventually stays above $M$, almost surely. Intersect this probability-one assertion over integers $M$. The result is $X_t\to\infty$. If zero is ever reached it remains occupied forever, and for initial state zero this is immediate. \b[Thus $\boxed{\mathbb P_i(E\cup F)=1}$ for every initial state.]
This uses the usual assumption implicit in an $I$-valued chain defined for all $t\geq0$: no explosion with loss to a cemetery state. The rate conditions alone do not ensure it. For example rates $q_{i,i+1}=i^3$, $q_{i0}=1$ allow an infinite upward path with probability $\prod_i i^3/(i^3+1)>0$ and a finite total holding time, since $\sum_i1/(i^3+1)<\infty$. The minimal chain then explodes; with a cemetery state the printed dichotomy would require an additional explosion alternative.
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