Answer
Amplitude $=\dfrac{5}{3}$
Period$=3$
Work Step by Step
Recall:
For $y=A\sin{(ωx)}$, we have:
$\text{Amplitude}=|A|\quad \quad \text{and} \quad \quad \text{Period}=T=\frac{2\pi}{ω}$
Using the fact that $\sin(-\theta)=-\sin\theta$, then
$$y=\frac{5}{3}\sin \left(-\frac{2\pi}{3}x\right)=-\frac{5}{3}\sin{\left(\frac{2\pi}{3}x\right)}$$
(By writing like this, we assured that $\omega$ is positive)
In $y=-\frac{5}{3}\sin \left(\frac{2\pi}{3}x\right)$ we have $A=-\frac{5}{3}$ and $\omega=\frac{2\pi}{3}$, so
Amplitude= $|A|=|\frac{5}{3}|=\frac{5}{3}$
Period $=T=\frac{2\pi}{\omega}=\dfrac{2\pi}{\frac{2\pi}{3}}=3$
