If $a$ and $b$ satisfy the equations $a +\frac1b=4$ and $\frac1a+b=\frac{16}{15}$, determine the product of all possible values of $ab$.

1. Given the equations:
   \[
   a + \frac{1}{b} = 4
   \]
   \[
   \frac{1}{a} + b = \frac{16}{15}
   \]

2. Multiply the two equations together:
   \[
   \left(a + \frac{1}{b}\right) \left(\frac{1}{a} + b\right) = 4 \cdot \frac{16}{15}
   \]
   Simplify the right-hand side:
   \[
   4 \cdot \frac{16}{15} = \frac{64}{15}
   \]

3. Expand the left-hand side:
   \[
   \left(a + \frac{1}{b}\right) \left(\frac{1}{a} + b\right) = a \cdot \frac{1}{a} + a \cdot b + \frac{1}{b} \cdot \frac{1}{a} + \frac{1}{b} \cdot b
   \]
   Simplify each term:
   \[
   1 + ab + \frac{1}{ab} + 1 = \frac{64}{15}
   \]
   Combine like terms:
   \[
   2 + ab + \frac{1}{ab} = \frac{64}{15}
   \]

4. Subtract 2 from both sides:
   \[
   ab + \frac{1}{ab} = \frac{64}{15} - 2
   \]
   Simplify the right-hand side:
   \[
   \frac{64}{15} - 2 = \frac{64}{15} - \frac{30}{15} = \frac{34}{15}
   \]
   Thus:
   \[
   ab + \frac{1}{ab} = \frac{34}{15}
   \]

5. Let \( x = ab \). Then we have:
   \[
   x + \frac{1}{x} = \frac{34}{15}
   \]

6. Multiply both sides by \( x \) to obtain a quadratic equation:
   \[
   x^2 + 1 = \frac{34}{15} x
   \]
   Rearrange to standard quadratic form:
   \[
   15x^2 - 34x + 15 = 0
   \]

7. Use the quadratic formula to solve for \( x \):
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Here, \( a = 15 \), \( b = -34 \), and \( c = 15 \):
   \[
   x = \frac{34 \pm \sqrt{(-34)^2 - 4 \cdot 15 \cdot 15}}{2 \cdot 15}
   \]
   Simplify under the square root:
   \[
   x = \frac{34 \pm \sqrt{1156 - 900}}{30}
   \]
   \[
   x = \frac{34 \pm \sqrt{256}}{30}
   \]
   \[
   x = \frac{34 \pm 16}{30}
   \]
   Thus, the two solutions are:
   \[
   x = \frac{34 + 16}{30} = \frac{50}{30} = \frac{5}{3}
   \]
   \[
   x = \frac{34 - 16}{30} = \frac{18}{30} = \frac{3}{5}
   \]

8. The product of all possible values of \( ab \) is:
   \[
   \left(\frac{5}{3}\right) \left(\frac{3}{5}\right) = 1
   \]

The final answer is \(\boxed{1}\).