Решение:

1. a) $$\frac{a+3}{a-1} - \frac{a}{1-a} = \frac{a+3}{a-1} + \frac{a}{a-1} = \frac{a+3+a}{a-1} = \frac{2a+3}{a-1}$$
2. б) $$\frac{3x+2y}{2x-3y} - \frac{x-8y}{3y-2x} = \frac{3x+2y}{2x-3y} + \frac{x-8y}{2x-3y} = \frac{3x+2y+x-8y}{2x-3y} = \frac{4x-6y}{2x-3y} = \frac{2(2x-3y)}{2x-3y} = 2$$
3. в) $$\frac{b^2}{2b-10} + \frac{25}{10-2b} = \frac{b^2}{2b-10} - \frac{25}{2b-10} = \frac{b^2-25}{2b-10} = \frac{(b-5)(b+5)}{2(b-5)} = \frac{b+5}{2}$$