a) $\{1, \frac{1}{2}, \frac{1}{4}, \frac{1}{16}\}$ is an example of a powerful set with four elements. (Part b shows that this is the unique powerful set with four elements.)

b) First we prove a lemma.

Lemma 1. A finite powerful set $S$ can not have an element greater than one and an element less than one.

Proof. Suppose by contradiction that there exist such elements. Now let $a$ be the least element of $S$ and $b$ the least element of $S$ which is greater than $1$. By assumption, we must have $a < 1$. Now $a < 1 < b$ and so

* $a^b < a^1 = 1$, but $a$ was the least element of $S$. Thus, $a^b \notin S$.
* $1 < b^a < b^1 = b$, but $b$ was the least element of $S$ greater than one. Therefore, $b^a \notin S$.

So our assumption leads to a contradiction proving the lemma. □

According to this lemma, if $S$ is a finite powerful set, all of its elements are in $[1, \infty)$ or in $(0, 1]$.

Firstly, suppose that $S$ is a powerful set with $n > 3$ number of elements in $[1, \infty)$ ($S = \{1 = a_1 < a_2 < \cdots < a_n\}$). Note that we can assume that $a_1 = 1$, because if $1 \notin S$, we can add it to $S$ to get a powerful set with more elements. For $i \ge 2$, $a_n^{a_i} > a_n$ and so $a_i^{a_n}$ must be in $S$. We have
$$a_1 < a_2 < a_2^{a_n} < a_3^{a_n} < \cdots < a_{n-1}^{a_n}$$
So for $2 \le i \le n - 1$, $a_i^{a_n} = a_{i+1}$. Now for $a_2 < a_{n-1}$ ($n > 3$) we have
$$a_2 < a_2^{a_{n-1}} < a_2^{a_n} = a_3 \Rightarrow a_2^{a_{n-1}} \notin S$$
$$a_{n-1} < a_{n-1}^{a_2} < a_{n-1}^{a_n} = a_n \Rightarrow a_{n-1}^{a_2} \notin S$$
but this contradicts, because $S$ was a powerful set.

Now suppose that $S$ is a powerful set with $n > 4$ elements in $(0, 1]$. Let $S = \{a_1 < a_2 < \cdots < a_n = 1\}$. Again we may assume $1 \in S$. Similar to the previous part, for each $1 \le i \le n - 2$, $a_{n-1} < a_{n-1}^{a_i} < 1$. So $a_n^{a_i} \notin S$, and consequently $a_i^{a_n} \in S$. We have
$$a_1 < a_1^{a_{n-1}} < a_2^{a_{n-1}} < \cdots < a_{n-2}^{a_n} < 1$$
So we get $a_i^{a_{n-1}} = a_{i+1}$ for each $2 \le i \le n - 2$. Now if we denote $a_{n-1}$ by $a$, we get
$$a_{n-2} = a^{1/a},\ a_{n-3} = a^{1/a^2}, \dots$$
Now by looking at $a_{n-1}$ and $a_{n-2}$, we conclude
$$a_{n-1} = a_{n-2}^{a_{n-1}} < a_{n-1}^{a_{n-2}} < 1 \Rightarrow a_{n-1}^{a_{n-2}} \notin S$$
This implies $a_{n-2}^{a_{n-1}} \in S$. Since $a_{n-2}^{a_{n-1}} > a_{n-2}^{a_n} = a_{n-1}$, we get $a_{n-2}^{a_{n-1}} = a_n$. So
$$a_{n-2}^{a_{n-1}} = (a^{1/a^2})^{(a^{1/a})} = a \implies a^{(a^{1/a})-2} = a \Rightarrow a^{1/a-2} = 1$$
But $a \ne 1$ and so $a = \frac{1}{2}$. Therefore, $a_{n-1} = \frac{1}{2}$, $a_{n-2} = \frac{1}{4}$ and $a_{n-3} = \frac{1}{16}$. Now since $n > 4$, $a_{n-4} = \frac{1}{256} \in S$ but it is easy to see none of $a_{n-3}^{a_{n-4}}$ and $a_{n-4}^{a_{n-3}}$ is in $S$. So there is no powerful set with more than 4 elements.