Answer
$\frac{\sqrt3-1}{2}.
$
Work Step by Step
We have the integral
$$
\int_{0}^{\pi/3} \int_{0}^{\pi/6} \sin (x+y) d x d y=\int_{0}^{\pi/3} (- \cos(x+y) )_{0}^{\pi/6} d y\\
=\int_{0}^{\pi/3} - \cos( y+\pi/6 )+ \cos yd y\\
=- \sin( y+\pi/6 )+ \sin y|_{0}^{\pi/3}=\frac{\sqrt 3}{2}-1+\frac{1}{2}=\frac{\sqrt3-1}{2}.
$$
