(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.
The only way to get a binomial distribution is to assume that the sampling is done with replacement. If the sampling is without replacement, the distribution of the number of defectives is hypergeometric.
For your second question, if you pick $5$ items and number them, the probability the first is defective is $\frac{10}{25}$. The probability the second is defective is also $\frac{10}{25}$. And so on. Many people find this unintuitive at first.
Of course the conditional probability the second is defective, given the first is, is not $\frac{10}{25}$. When we do sampling without replacement, we lose independence.
Best Answer
Let me take a concrete example, to illustrate the way to solve.
Total $= n = 10$, defectives $= r = 3$, place where last defective found $=k = 6$
We need $(3-1) = 2$ defectives in the first $5$ followed by the final defective, against ${10\choose3}$ random places for the defectives, thus
$$Pr = \frac{{5\choose2}{1\choose1}}{10\choose3}$$
You should now be able to convert this to a general formula using $n,k,r$
[ Of course, ${1\choose1}$ can be omitted, was included to give better understanding ]