(a) Let $G$ be the event that the whole lot is good (zero defectives), and let $P$ be the event there were no defectives in the sample of $2$. We are asked for the conditional probability $\Pr(G|P)$, the probability that the whole lot is good given the information that the sample of $2$ had no defectives. By a formula which is likely familiar to you, essentially the definition of conditional probability, we have $\Pr(G|P)=\frac{\Pr(G\cap P}{\Pr(P)}.$ It remains to find the probabilities on the right.
We go first for the harder one, $\Pr(P)$. The event $P$ can happen in three ways: (i) the chosen lot has no defectives, and (of course) no defectives are found; (ii) the chosen lot has one defective, and no defectives are found; or (iii) the chosen lot has two defectives, and no defectives are found.
The probability of (i) is clearly $0.6$. Note that this is also $\Pr(P\cap G)$.
The probability that the chosen lot has $1$ defective is $0.3$. If we sample from this lot, then with probability $\frac{19}{20}$ the first tested item is OK, and given that happened, the probability the second item tested is OK is $\frac{18}{19}$, for a product of $\frac{18}{20}$. So the probability of (ii) is $(0.3)(18/20)$, which is exactly $0.27$.
In the same way, we find that the probability of (iii) is $(0.1)(18/20)(17/19)$, which is approximately $0.0805$.
So $\Pr(P)\approx 0.6+0.27+0.0805$, which is about $0.9505$.
Finally, for our conditional probability, divide $\Pr(G\cap P)$ by $\Pr(P)$. We get about $0.63123$.
(b) We are expected to assume independence. For the inspected lot, the probability it has no defectives is approximately $0.63213$. The $9$ uninspected lots each have probability $0.6$ of having no defectives. Multiply.
Remark: Our $0.63123$ is a little bigger than your conjectured answer of $0.6$. This is because the fact that no bads were found in our inadequate inspection makes it somewhat more likely that the lot is an all-good lot.