, 2016

Solution:

We multiply $a + \frac{1}{b} = 4$ and $\frac{1}{a} + b = \frac{16}{15}$ to get
$$\left(a + \frac{1}{b}\right) \left(\frac{1}{a} + b\right) = 4 \cdot \frac{16}{15} = \frac{64}{15}.$$
Expanding the left side:
$$a \cdot \frac{1}{a} + a b + \frac{1}{b} \cdot \frac{1}{a} + \frac{1}{b} b = 1 + a b + \frac{1}{a b} + 1 = 2 + a b + \frac{1}{a b}.$$
So,
$$2 + a b + \frac{1}{a b} = \frac{64}{15}.$$
Subtract $2$ from both sides:
$$a b + \frac{1}{a b} = \frac{64}{15} - 2 = \frac{64 - 30}{15} = \frac{34}{15}.$$
Let $x = a b$. Then
$$x + \frac{1}{x} = \frac{34}{15}.$$
Multiply both sides by $x$:
$$x^2 - \frac{34}{15} x + 1 = 0.$$
By Vieta's formulas, the product of the roots is $1$.

Therefore, the product of all possible values of $a b$ is $1$.