Let today, tomorrow and the day after tomorrow correspond with $0,1,2$ respectively.
Then in your work on a) you actually calculate $\Pr(S_0\cap C_2)$ under your own (hence questionable) assumption that $\Pr(S_0)=0.7$.
However, you are asked to calculate $\Pr(C_2\mid S_0)$.
You make the same mistake in b).
Dividing both original answers by $\Pr(S_0)$ will repair.
This comes to the same as striping away the first 0.7 (as TonyK) suggests in his comment.