For the second part, it is probably easiest to proceed recursively.
We are told $P(A_1)=.2$.
Now suppose we have computed $P(A_{j-1})$ and want to compute $P(A_j)$. Look at the first $j-1$ shots. We only care about the last one, which we know to be a hit with probability $P(A_{j-1})$ (and therefore a miss with probability $ 1-P(A_{j-1}))$ It follows that $$P(A_j)=.6\times P(A_{j-1})+ .3\times (1-P(A_{j-1}))=.3 \times P(A_{j-1})+.3 = .3\times (P(A_{j-1})+1)$$
In this way it is very easy to compute all the $P(A_j)$.
Note: it wasn't part of the question, but it is interesting to note that in the limit as the number of free throws gets large, the probability goes to $\frac 37$; this follows quickly from the recursion. Indeed, the recursion can be solved in closed form by standard methods. Doing so, we get: $$P(A_j)=\frac 37-\frac {16}{21}(.3)^j$$
The answer to the first question is wrong.
To see this, suppose that the probability of success at each shot is $0.5$. That means that you can work out the probability without any calculation: just count among the $2\times 2\times 2\times 2\times 2=32$ possible results.
The probability of getting at least 3 successes is $\frac {16}{32}$.
The probability of getting precisely 3 successes is $\frac{10}{32}$.
Neither of these figures matches what the method quoted for the first example would give, which is $0.5^3\times 0.5^2=\frac 1 {32}$.
I have changed the probability to $0.5$ to show that the method given produces wrong results, in a case where it easily to see without calculation what the right result should be. A method which fails for $p=0.5$ will also fail for $p=0.8$ - its failure will be less easy to see, that’s all.
Since the method given is nonsensical, no wonder you are confused!
The second example is simple to understand if you are not confused by the first one. There is no third drawing. The probability of success on two drawings is simply the square of the probability of success on one.
EDIT: As drhab points out, the answer that is being given in the first example is to the question “What is the probability of three successes followed by two failures?” His solution of the question as you stated it is correct and he should make it into a proper answer instead of a comment.
Best Answer
I will work out the probability of making the $(n+1)$'th throw, given that they made the $n$'th thow. Let $H_n$ be the event that they hit on the $n$'th shot, and let $K_n$ be the total number of hits after $n$ shots.
I will first prove by induction, that after $n$ throws there is an equal probability of having any number of hits. I.e., we have $P(K_n=k) = \frac{1}{n-1}$ for all $1\le k \le n-1$.
The case $n=2$ is trivial. Assume $n\ge3$. We can get $K_{n+1}=k$ in two ways: Hitting after $k-1$ hits, or missing after $k$ hits. This means: $$ P(K_{n+1}=k) = \frac{1}{n-1} \cdot \frac{k-1}{n} + \frac{1}{n-1} \cdot \frac{n-k}{n} = \frac{1}{n} $$
Note that this works even for the edge cases $k=1$ and $k=n$. Now we come back to the original problem. We want to work out: $$ P(H_{n+1}|H_n) = \frac{P(H_{n+1} \cap H_{n})}{P(H_{n})} $$
for $n\ge 3$. By the symmetry of the setup, we have simply $P(H_{n})=1/2$. We can calculate $P(H_{n+1} \cap H_{n})$ by splitting up in the cases for $K_{n-1}$: $$ \begin{split} P(H_{n+1} \cap H_{n}) &= \sum_{k=1}^{n-2} \frac{1}{n-2} \cdot \frac{k}{n-1}\cdot\frac{k+1}{n} = \frac{1}{n(n-1)(n-2)} \left(\sum_{k=1}^{n-2} k(k+1)\right) \\ &= \frac{1}{n(n-1)(n-2)} \cdot \frac{n(n-1)(n-2)}{3} = \frac13 \end{split} $$
So in the end we get $$ P(H_{n+1}|H_n) = \frac{1/3}{1/2} = \frac23 $$
Note that the result is independent of $n$!
Edit:
On antkam's suggestion, I'll prove my observation in the comments. I claim that all sequences of $n$ shots that have the same number of hits are equally likely. (This can actually be proven from the property I proved inductively above, but I'll do it the other way round). For example $P(HHMM) = P(HMMH) = P(MHMH) = \ldots$, where $H$ is a hit and $M$ is a miss. Since the first two shots are fixed, the sequences begin at the third shot. This is interesting, because while there are less ways to get a very high or low number of hits, each of those sequences are more likely because of the setup. These tendencies exactly cancel out to give the uniform distribution of $K_n$.
My precise claim is this: $$ P(S_3S_4\cdots S_{n+1}) = \frac{k!(n-k-1)!}{n!} $$ for any $n+1 \ge 3$, where $S_i$ is the outcome of the $i$'th shot (hit or miss), and $k$ is the total number hits.
Let $p_n = P(S_n = H)$ and $q_n = P(S_n = M)$. Note that we can write $$ P(S_3S_4\cdots S_{n+1}) = \prod_{S_i=H}p_i \cdot \prod_{S_i=M}q_i $$ We have $p_i=\frac{K_i}{i-1}$ and $q_i=\frac{i-1-K_i}{i-1}$. Since the sequence goes from the third to the $(n+1)$'th shot, we get a denominator of $n!$ when we multiply all the probabilities.
Let's consider the numerator. If we hit on the $i$'th shot ($S_i = H$), then $$ p_{i+1} = \frac{K_i+1}{i}, \quad q_{i+1}=\frac{i-(K_i+1)}{i} $$
and if $S_i = M$: $$ p_{i+1} = \frac{K_i}{i}, \quad q_{i+1}=\frac{i-K_i}{i} $$
We see that when we hit, the numerator of $p$ goes up by 1, and when we miss, the numerator of $q$ goes up by 1. Meanwhile the other numerators are unchanged. Note that $p_3 = q_3 = 1/2$, so they both start with a numerator of 1. Thus, multiplying all the numerators together, we get $k!(n-k-1)!$, which proves the claim.
So how does this get us the uniform distribution of $K_n$? If $K_{n+1} = k$, then the sequence of $n-1$ shots will have $k$ hits. There are $\binom{n-1}{k}$ such sequences, all equally likely, so we rediscover the result that: $$ P(K_{n+1}=k) = \frac{k!(n-k-1)!}{n!}\cdot \binom{n-1}{k} = \frac{k!(n-k-1)!}{n!} \cdot \frac{(n-1)!}{k!(n-k-1)!} = \frac{1}{n} $$