Answer
$\frac{1}{3}$; $\frac{2\pi}{3}$
Work Step by Step
We know that if $A$ is a real number and $B\gt0$, then the graphs of $y=Asin(Bx)$ and $y=Acos(Bx)$ will have an amplitude of $|A|$ and a period of $\frac{2\pi}{B}$.
Therefore, the amplitude of the graph of $y=\frac{1}{3}sin3x$ is $|\frac{1}{3}|=\frac{1}{3}$
Therefore, the period of the graph of $y=\frac{1}{3}sin3x$ is $\frac{2\pi}{3}$.