a)
$$\frac{3-2a}{2a} - \frac{1-a^2}{a} = \frac{3-2a}{2a} - \frac{2(1-a^2)}{2a} = \frac{3-2a - 2 + 2a^2}{2a} = \frac{2a^2 - 2a + 1}{2a}$$
б)
$$\frac{1}{3x+y} - \frac{1}{3x-y} = \frac{3x-y}{(3x+y)(3x-y)} - \frac{3x+y}{(3x+y)(3x-y)} = \frac{3x-y - (3x+y)}{(3x+y)(3x-y)} = \frac{3x-y-3x-y}{9x^2 - y^2} = \frac{-2y}{9x^2 - y^2} = \frac{2y}{y^2 - 9x^2}$$
Ответ: a) $$\frac{2a^2 - 2a + 1}{2a}$$, б) $$\frac{2y}{y^2 - 9x^2}$$