Hi guys, preparing for my finals and trying to get this question out for practice. The exam is in a couple of hours so apologies for being brief.
I think I have computed parts 1 and 2 fine.
$$0.3*0.1 + 0.7*0.3 = 0.24$$
for the first part, and
$$P(\text{parasite} \mid A) = \frac{P(A \mid \text{parasite})P(\text{parasite})}{P(A)}$$
$$= \frac{0.7*0.3}{0.3*0.7 + 0.7*0.2} = 0.6$$
Hopefully these parts are both correct so far.
Now for part iii), I started out $$P(A_2 \mid A_1) = \frac{P(A_1 \mid A_2)P(A_2)}{P(A_1)}$$ but this doesn't really make any sense. So then I tried
$$P(A_2 \mid A_1)= P(A_2 \mid A_1, \text{parasite}) + P(A_2 \mid A_1, \text{no parasite}) $$
I'm not sure where to go with it after this or if that approach is correct at all. I know I ought to look for a hint and then work through the rest myself but as the exam is so soon I would really appreciate someone pointing out the next couple of steps, and maybe even a hint/setup for part iv).
Your second approach is the right idea, but what you were after is:
$$\mathsf P(A_2\mid A_1) ~=~ \mathsf P(A_2\mid P_{\rm arasite})~\mathsf P(P_{\rm arasite}\mid A_1)+\mathsf P(A_2\mid P_{\rm arasite}^\complement)~\mathsf P(P_{\rm arasite}^\complement\mid A_1)$$
Now you are looking for $\mathsf P(P_{\rm arasite}\mid A_1,A_2)$ using Bayes' Rule, and the fact that the events of the two butterflies both being type A are conditionally independent given the presence/absence of the parasite.