We have $a_{n+1} - 1 = a_n(a_n-1)$. Part (a). Note that \[ \frac{a_{n+1}-1}{a_n-1} = a_n\implies a_m =1+a_1a_2\cdots a_{m-1}. \]Hence if $m>n$ then $a_n\mid a_m-1$, yielding coprimality. Part (b). Now we have \[ \frac{1}{a_n-1}-\frac{1}{a_n} = \frac{1}{a_{n+1}-1} \implies\sum_{1\le k\le n}\frac{1}{a_k} = 1-\frac{1}{a_{n+1}-1}. \]Now clearly $a_n\to\infty$ as $n\to\infty$. From here, the conclusion follows. Remark. A reheated version of this problem appeared in 2003 Romanian TST.

It is easy to show by induction that $a_{n+1} = a_1a_2 \dots a_n + 1$, thus (a) is clear. For (b), let $S_n = \sum_{k=1}^{n} \frac{1}{a_k}$. We will prove by induction that $S_n = \frac{a_{n+1} - 2}{a_{n+1} - 1}$. For $n=1$ it is clear. Assume true for $n$. We have $S_{n+1} = S_n + \frac{1}{a_{n+1}} = \frac{a_{n+1}^2 - a_{n+1} - 1}{a_{n+1}^2 - a_{n+1}} = \frac{a_{n+2} - 2}{a_{n+2} - 1}$, thus we are done by induction. So, $\lim_{n\to\infty}\sum_{k=1}^n\frac1{a_k} = \lim_{n\to\infty} S_n = \lim_{n\to\infty} \frac{a_{n+1} - 2}{a_{n+1} - 1} = 1$, as clearly $a_n$ is unbounded.