$Ans: 3$
1. $At\ x=0 \sin x=0$ & $\frac{x}{3}=0 $
2. $At\ x=\frac{\Pi}{2},\ \sin x=1\ but\ \frac{x}{3}=\frac{\pi}{6}\ i.e\ \sin x > \frac{x}{3} $ &
$At\ x=\Pi,\ \sin x=0\ but\ \frac{x}{3}=\frac{\pi}{3}\ i.e\ \sin x < \frac{x}{3} $
So according to $Intermediate\ Value\ Theorem$ there exists atleast one solution in the range $\left[\frac{\Pi}{2},\pi \right]$
And from $x=\frac{\Pi}{2}\ to\ x=\pi,\ \sin x$ is decreasing and $\frac{x}{3}$ is increasing, so there will be only one solution in the range $\left[\frac{\Pi}{2},\pi \right]$
3. Similarly one solution exists in the range $\left[\frac{\Pi}{2},\pi \right]$
P.S : $Intermediate\ Value\ Theorem$ : $Suppose\ f\ is\ function\ continuous\ at\ every\ point\ of\ the\ interval\ [a,b]\ :\ $

$f\ will\ take\ on\ every\ value\ b/w\ f(a)\ and\ f(b)\ over\ the\ interval$

$for\ any\ L\ b/w\ the\ values\ f(a)\ and\ f(b),\ there\ exist\ a\ number\ c\ in\ [a,b]\ for\ which\ f(c)=L\ $