Answer
The amplitude is $3$, the period is $4$, there is no vertical translation and the phase shift is $\frac{1}{2}$ units to the right since $d$ is more than zero.
Work Step by Step
We first write the equation in the form $y=c+a \cos [b(x-d)]$. Therefore, $y=3\cos [\frac{\pi}{2}(x-\frac{1}{2})$ becomes $y=0+3\cos [\frac{\pi}{2}(x-\frac{1}{2})$
Comparing the two equations, $a=3,b=\frac{\pi}{2},c=0$ and $d=\frac{1}{2}$.
The amplitude is $|a|=|3|=3.$
The period is $\frac{2\pi}{b}=\frac{2\pi}{0.5\pi}=4$.
The vertical translation is $c=0$.
The phase shift is $|d|=|\frac{1}{2}|=\frac{1}{2}$
Therefore, the amplitude is $3$, the period is $4$, there is no vertical translation and the phase shift is $\frac{1}{2}$ units to the right since $d$ is more than zero.