jmc

There's a $\frac{1}{3}$ chance that I will select each team. Once I have selected a team, let $n$ be the number of students on that team. There are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ ways to choose a pair of those students to give books to, but only one of those pairs will be the two co-captains, which means that once I have selected that team, the probability that I give books to the co-captains is $$\frac{1}{\frac{n(n-1)}{2}}=\frac{2}{n(n-1)}.$$Since the teams have $5,$ $7,$ and $8$ students, this means that the total probability is $$\frac{1}{3}\left(\frac{2}{5(5-1)}+\frac{2}{7(7-1)}+\frac{2}{8(8-1)}\right)$$which after a bit of arithmetic simplifies to $\boxed{\frac{11}{180}}$.