Answer
$Amplitude = \frac{2}{3}$
$Period = \frac{2\pi}{3}$
$Horizontal\ shift = -\frac{\pi}{6}$ i.e left shift $\frac{\pi}{6}$
$Vertical\ shift = 0$ (No vertical shift)
Work Step by Step
If C is any real number and $B> 0$, then the graphs of $y = k + A\sin(Bx+C)$ and $y = k + A\cos (Bx+C)$ will have
$Amplitude = |A|$
$Period = \frac{2\pi}{B}$
$Horizontal\ shift = –\frac{C}{B}$
$Vertical\ shift = k$
so for $y = -\frac{2}{3}\sin (3x + \frac{\pi}{2} )$
$Amplitude = |-\frac{2}{3}| = \frac{2}{3}$
$Period = \frac{2\pi}{3}$
$Horizontal\ shift = (–\frac{\frac{\pi}{2}}{3}) = -\frac{\pi}{6}$ i.e left shift $\frac{\pi}{6}$
$Vertical\ shift = 0$ (No vertical shift)