Haven't done question 1, but here's 2.
\frac{3+2isinx}{1-2isinx}
$Realising the denominator $ \frac{3+2isinx}{1-2isinx} \times \frac{1+2isinx}{1+2isinx}
=\frac{3-4sin^2x + 8isinx}{1+4sin^2x}
$For it to be purely imaginary, $ 3-4sin^2x $ must equal 0$
3-4sin^2x=0
sinx=\pm\frac{\sqrt{3}}{2}...