An electric bulb, linked to a make and break power supply, turns...

The proper time interval between two consecutive switching events in the rest frame of the bulb is $\Delta t_0 = 1 \text{ s}$.

The spaceship is moving with a speed $v = 0.8c$ relative to the bulb.

According to the time dilation formula, the time interval $\Delta t$ observed from the spaceship is given by:

$\Delta t = \dfrac{\Delta t_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

Substituting $v = 0.8c$ and $\Delta t_0 = 1 \text{ s}$:

$\Delta t = \dfrac{1}{\sqrt{1 - (0.8)^2}} = \dfrac{1}{\sqrt{1 - 0.64}} = \dfrac{1}{\sqrt{0.36}} = \dfrac{1}{0.6} \text{ s}$

The observed frequency $f$ is the reciprocal of the observed time interval:

$f = \dfrac{1}{\Delta t} = 0.6 \text{ Hz}$