Answer
$Amplitude= \frac{1}{3} $
$Period = 2$
$Horizontal\ shift $= left shift $\frac{3}{2}$
$Vertical\ shift = 2$ upward 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 $y = 2 - \frac{1}{3}\cos (\pi x + \frac{3\pi}{2} )$
$Amplitude = |-\frac{1}{3}| = \frac{1}{3} $
$Period = \frac{2\pi}{\pi} = 2$
$Horizontal\ shift = (–\frac{\frac{3\pi}{2}}{\pi}) = -\frac{3}{2}$ i.e left shift $\frac{3}{2}$
$Vertical\ shift = 2$ ($2$ upward shift)