(b)

Given trigonometric relation is

$\left(\frac{1}{3} \sin \theta+\frac{2}{3} \cos \theta\right)^2=\frac{1}{3} \sin ^2 \theta+\frac{2}{3} \cos ^2 \theta$

$\Rightarrow \quad \frac{1}{9} \sin ^2 \theta+\frac{4}{9} \cos ^2 \theta+\frac{4}{9} \sin \theta \cos \theta$

$=\frac{1}{3} \sin ^2 \theta+\frac{2}{3} \cos ^2 \theta$

$\Rightarrow \quad \frac{2}{9} \sin ^2 \theta+\frac{2}{9} \cos ^2 \theta-\frac{4}{9} \sin \theta \cos \theta=0$

$\Rightarrow \quad \sin ^2 \theta+\cos ^2 \theta-2 \sin \theta \cos \theta=0$

$\Rightarrow \quad \quad \sin 2 \theta=1$

$\Rightarrow \quad 2 \theta=2 n \pi+\frac{\pi}{2}, n \in I$

$\Rightarrow \quad \theta=n \pi+\frac{\pi}{4}, n \in I$

$\therefore \quad A=\left\{\theta \in R: \theta=n \pi+\frac{\pi}{4}, n \in I\right\}$

$\therefore A \cap[0, \pi]=\left\{\frac{\pi}{4}\right\}$

$\therefore \quad A \cap[0, \pi]$ has exactly one point.