Addition and substraction of algebraic fractions -urgent questions

A very good day to everyone here on Math Help Forum!

I have a few questions related to algebraic fractions. They are as the following:

1. $\displaystyle \dfrac{x^2}{y}$ $\displaystyle + \dfrac{y}{x^2}$

2. $\displaystyle \dfrac{x}{x+16} $$\displaystyle +$$\displaystyle \dfrac{5}{x^2 +x-12}$

3. $\displaystyle \dfrac{5}{2x^2 +5x +3}$$\displaystyle - \dfrac{1}{2x^2 +3x -2}$

My workings are shown below:

1. $\displaystyle \dfrac{x^2}{y}$ $\displaystyle + \dfrac{y}{x^2}$

$\displaystyle = \dfrac{x^2y+x^2y}{x^2y}$

$\displaystyle =\dfrac{2x^2y}{x^2y}$

$\displaystyle

=2

$ ** ?**

2. $\displaystyle \dfrac{x}{x+16} $$\displaystyle +$$\displaystyle \dfrac{5}{x^2 +x -12}$

$\displaystyle =\dfrac{x}{x^2 -4^2}$ $\displaystyle + \dfrac{5}{(x+4)(x-3)}$

$\displaystyle =\dfrac{x}{(x-4)(x+4)} $$\displaystyle + \dfrac{5}{(x+4)(x-3)}$

$\displaystyle

=\dfrac{x(x-3)+5(x-4)}{(x-4)(x+4)(x-3)}

$

$\displaystyle

=\dfrac{x^2+2x-20}{(x-4)(x+4)(x-3)}

$

$\displaystyle

=\dfrac{(x+6)(x-4)}{(x-4)(x+4)(x-3)}

$

$\displaystyle

= \dfrac{x+6}{(x+4)(x-3)}

$ ** ?**

3.$\displaystyle \dfrac{5}{2x^2 +5x +3}$$\displaystyle - \dfrac{1}{2x^2 +3x -2}$

$\displaystyle =\dfrac{5}{(2x+3)(x+1)}$ $\displaystyle - \dfrac{1}{(2x-1)(x+2)}$

$\displaystyle

=\dfrac{5(2x-1)(x+2) -1(2x+3)(x+2)}{(2x+3)(x+1)(2x-1)(x+2)}

$

*This is where I am stuck.*

If you happened to notice any mistakes and blunders, please pinpoint them and explain how it should be done.

Thank you so much for taking your time!