Решение:

1) a) $$\frac{y}{4} + \frac{y-2}{5} = \frac{5y + 4(y-2)}{20} = \frac{5y + 4y - 8}{20} = \frac{9y - 8}{20}$$

г) $$\frac{c+3}{c^2} - \frac{1}{c} = \frac{c+3 - c}{c^2} = \frac{3}{c^2}$$

2) a) $$\frac{(a - b)^2}{18b} - \frac{(a - b)^2}{12b} = \frac{2(a - b)^2 - 3(a - b)^2}{36b} = \frac{-(a - b)^2}{36b} = -\frac{(a - b)^2}{36b}$$

3) a) $$\frac{c-2}{3(c + 4)} + \frac{c}{c + 4} = \frac{c-2 + 3c}{3(c + 4)} = \frac{4c - 2}{3(c + 4)} = \frac{2(2c - 1)}{3(c + 4)}$$

4) a) $$\frac{x + 4}{xy - x^2} + \frac{y + 4}{xy - y^2} = \frac{x + 4}{x(y - x)} + \frac{y + 4}{y(x - y)} = \frac{y(x + 4) - x(y + 4)}{xy(y - x)} = \frac{xy + 4y - xy - 4x}{xy(y - x)} = \frac{4(y - x)}{xy(y - x)} = \frac{4}{xy}$$

в) $$\frac{4}{c^2-9} - \frac{2}{c^2 + 3c} = \frac{4}{(c-3)(c+3)} - \frac{2}{c(c+3)} = \frac{4c - 2(c-3)}{c(c-3)(c+3)} = \frac{4c - 2c + 6}{c(c-3)(c+3)} = \frac{2c + 6}{c(c-3)(c+3)} = \frac{2(c + 3)}{c(c-3)(c+3)} = \frac{2}{c(c-3)}$$