Align clock times before comparing normal travel times

$P(X>t)=0.9$ means $P(X<t)=0.1$. Since $X\sim N(30,9)$, $t=30+3\Phi^{-1}(0.1)=26.1553\ldots\approx26.2$ minutes, to 3 s.f.

Midnight is 31 minutes after the train departs and 20 minutes after Abel starts. Using the independence assumed in the printed working, $P(X>31,Y<20)=P(X>31)P(Y<20)\approx0.361$.

The departure times differ by 11 minutes. Abel arrives first when $11+Y<X$, or $X-Y>11$. Under independence, $X-Y\sim N(30-16,9+4)=N(14,13)$, so $P(X-Y>11)\approx0.797$.

The event in(ii) is contained in the catch-the-train event, but the latter also allows both arrivals before midnight or both after midnight with Abel first. The source gives a train after 12.01am with Abel before 12.01am as an example; cases where Abel arrives between midnight and 12.01am add possibilities absent from (ii). These extra events have positive probability.

A normal $N(4,6)$ would put about 99.7% of its values within $4\pm3\sqrt6$, a range extending well below zero. More directly, $P(W<0)=\Phi(-4/\sqrt6)\approx0.0512$. Roughly 5.12% impossible negative times makes this a poor normal model for the individual walking time.

For independent identically distributed trips, $\bar W\mathrel{\dot\sim}N(4,6/40)$ by the CLT. Therefore $P(\bar W<3.5)\approx\Phi[-0.5/\sqrt{6/40}]=0.0984$, to 3 s.f.