Answer
$\frac{1}{3}$
Work Step by Step
\begin{align*}
\lim _{z\rightarrow 0} \frac{\sin (z/3)}{\sin z} &=\lim _{z\rightarrow 0} \frac{\sin (z/3)}{z} \frac{z}{\sin z} \\
&= \lim _{z\rightarrow 0} \frac{1}{3} \frac{\sin (z/3)}{z/3} \frac{z}{\sin z} \\
&=\frac{1}{3}\lim _{z/3\rightarrow 0} \frac{\sin (z/3)}{z/3}\lim _{z\rightarrow 0} \frac{z}{\sin z} \\
&=\frac{1}{3}.
\end{align*}
Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1. $
