а) $$\frac{2a-1}{3a} + \frac{a+2}{6a} = \frac{2(2a-1)}{6a} + \frac{a+2}{6a} = \frac{4a-2+a+2}{6a} = \frac{5a}{6a} = \frac{5}{6}$$
б) $$\frac{1}{x-3} - \frac{1}{x+3} = \frac{(x+3) - (x-3)}{(x-3)(x+3)} = \frac{x+3-x+3}{x^2-9} = \frac{6}{x^2-9}$$