We first note that if $f$ is a semi-reciprocal function and $f(1) = a \in S$, then $f(a) = f(f(1)) = 1$. Hence, $a = f(1) = f(f(a)) = 1/a$, and so $a^2 = 1$. Since $S$ does not contain negative numbers, we must have $a = 1$, i.e. $f(1) = 1$ for each semi-reciprocal function $f$.

If $a \in S$ and $b = f(a)$, the condition $f(f(x)) = 1/x$ implies
$$f(b) = f(f(a)) = 1/a, \quad f(1/a) = f(f(b)) = 1/b, \quad f(1/b) = f(f(1/a)) = a,$$
i.e., the function $f$ cyclically permutes the four numbers $a, b, 1/a, 1/b$:
$$a \mapsto b \mapsto \frac{1}{a} \mapsto \frac{1}{b} \mapsto a. \qquad (5)$$

To solve (a) we construct a semi-reciprocal function $f$ by defining
$$\begin{aligned} f(1) &= 1 & f(2n) &= 2n + 1 & f(2n + 1) &= \frac{1}{2n} \\ f\left(\frac{1}{2n}\right) &= \frac{1}{2n + 1} & f\left(\frac{1}{2n + 1}\right) &= 2n. \end{aligned}$$
for all integers $n \ge 1$. This settles part (a). More generally, such functions can be obtained from a partition of the positive integers greater than 1 into ordered pairs $(a, b)$ by applying (5). In the example given above, the pairs $(2n, 2n + 1)$ are used.

To finish part (b), we first note that we have seen above that $f(p) = p$ has the solution $p = 1$. We need to show that there is no other solution.

Suppose $p \in S$ satisfies $f(p) = p$. Because $f$ is semi-reciprocal, we obtain
$$p = f(p) = f(f(p)) = 1/p,$$
hence $p^2 = 1$ and therefore $p = 1$, since $p \in S$ is positive.