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