a) $\frac{2+i}{3-2i}$

= $\frac{(2+i)(3+2i)}{(3-2i)(3+2i)}$

= $\frac{4+7i}{13}$

= $\frac{4}{13}+\frac{7i}{13}$

b) $\frac{1+i\sqrt{2}}{2+i\sqrt{3}}$

= $\frac{(1+i\sqrt{2})(2-i\sqrt{3})}{(2+i\sqrt{3})(2-i\sqrt{3})}$

= $\frac{2+\sqrt{6}}{7}+\frac{2\sqrt{2}-\sqrt{3}i}{7}$

c) $\frac{5i}{2-3i}$

= $\frac{5i(2+3i)}{(2-3i)(2+3i)}$

= $\frac{15+10i}{13}$

= $\frac{15}{13}+\frac{10i}{13}$

d) $\frac{5-2i}{i}$

= $(5-2i)(-i)$

= $-2-5i$