1. Interpret the situation

A fish is inside water and sees a bird in air. Because of refraction at the water surface, the fish does not see the bird at its real position. It sees an apparent position of the bird.

For an object in air viewed from water, the apparent height is reduced by the refractive index:

$h' = \frac{h}{\mu}$

where $\mu = \frac{4}{3}$.

2. Relate actual and apparent velocities of the bird

Let the actual downward speed of the bird be $v$. Then its apparent downward speed as seen by the fish will be

$v_{\text{app}} = \frac{v}{\mu}$

since apparent height is $1/\mu$ times the real height.

So,

$v_{\text{app}} = \frac{v}{4/3} = \frac{3v}{4}$

3. Use the observed relative speed

The fish rises upward with speed $8\ \text{m s}^{-1}$. The bird appears to the fish to be diving downward towards it with speed $12\ \text{m s}^{-1}$.

Thus, the relative speed of approach between fish and apparent bird is

$8 + v_{\text{app}} = 12$

because they move toward each other.

Substitute $v_{\text{app}} = \frac{3v}{4}$:

$8 + \frac{3v}{4} = 12$

$\frac{3v}{4} = 4$

$3v = 16$

$v = \frac{16}{3}\ \text{m s}^{-1}$

4. Final answer

The actual speed of the bird is

$\boxed{\frac{16}{3}\ \text{m s}^{-1}}$

Since this is an integer-type question, the numerical value is approximately

$\boxed{5.33}$

So the stored answer $3$ is not correct.