$a)\dfrac{1}{2+3i}=\dfrac{2-3i}{(2+3i)(2-3i)}=\dfrac{2-3i}{13}=\dfrac{2}{13}-\dfrac{3}{13}i\\ |z|=\sqrt{\dfrac{2}{13}^2+\dfrac{3}{13}^2}=\dfrac{\sqrt{13}}{13}\\ \overline{z}=\dfrac{2}{13}+\dfrac{3}{13}i\\ b)\dfrac{4+5i}{i}=\dfrac{(4+5i)i}{i^2}=-(4i-5)=5-4i\\ |z|=\sqrt{5^2+4^2}=\sqrt{41}\\ \overline{z}=5+4i$