$\displaystyle\frac{x+y}{xy}-\displaystyle\frac{x+1}{x}-\displaystyle\frac{y+1}{y}=$

$=\displaystyle\frac{x+y}{xy}-\displaystyle\frac{y(x+1)}{xy}-\displaystyle\frac{x(y+1)}{xy}=$

$=\displaystyle\frac{x+y-y(x+1)-x(y+1)}{xy}=$

$=\displaystyle\frac{x+y-xy-y-xy-x}{xy}=$

$=\displaystyle\frac{-2xy}{xy}= -2$ 

$\displaystyle\frac{2+\displaystyle\frac{3}{a}}{1+\displaystyle\frac{5}{a}}=$

$=\bigg(2+\displaystyle\frac{3}{a}\bigg):\bigg(1+\displaystyle\frac{5}{a}\bigg)=$

$=\bigg(\displaystyle\frac{2a}{a}+\displaystyle\frac{3}{a}\bigg):\bigg(\displaystyle\frac{a}{a}+\displaystyle\frac{5}{a}\bigg)=$

$=\displaystyle\frac{2a+3}{a}:\displaystyle\frac{a+5}{a}=$

$=\displaystyle\frac{2a+3}{a}\cdot\displaystyle\frac{a}{a+5}=$ 

$=\displaystyle\frac{2a+3}{a+5}$

$\displaystyle\frac{\displaystyle\frac{a+2}{a^2}+\displaystyle\frac{8}{a}}{3-\displaystyle\frac{1}{a}}=$

$=\bigg(\displaystyle\frac{a+2}{a^2}+\displaystyle\frac{8}{a}\bigg):\bigg(3-\displaystyle\frac{1}{a}\bigg)=$

$=\bigg(\displaystyle\frac{a+2}{a^2}+\displaystyle\frac{8a}{a^2}\bigg):\bigg(\displaystyle\frac{3a}{a}-\displaystyle\frac{1}{a}\bigg)=$

$=\displaystyle\frac{a+2+8a}{a^2}:\displaystyle\frac{3a-1}{a}=$

$=\displaystyle\frac{9a+2}{a^2}\cdot\displaystyle\frac{a}{3a-1}=$

$=\displaystyle\frac{9a+2}{3a^2-a}$