Answer
Amplitude $=\dfrac{9}{5}$
Period $=\dfrac{4}{3}$
Work Step by Step
Recall:
For $y=A\cos{(ωx)}$, we have:
$\text{Amplitude}=|A|\quad \quad \text{and} \quad \quad \text{Period}=T=\frac{2\pi}{ω}$
Using the fact that $\cos(-\theta)=\cos\theta$, then
$$y=\frac{9}{5}\cos{\left(-\frac{3\pi}{2}x\right)}=\frac{9}{5}\cos{\left(\frac{3\pi}{2}x\right)}$$
(This step is in order to make sure that $\omega$ is positive)
In $\frac{9}{5}\cos{\left(\frac{3\pi}{2}x\right)}$ we have $A=\frac{9}{5}$ and $\omega=\frac{3\pi}{2}$, so
Amplitude =$|A|=|\frac{9}{5}|=\frac{9}{5}$
Period $=T=\frac{2\pi}{\omega}=\dfrac{2\pi}{\frac{3\pi}{2}}=\dfrac{4}{3}$
