Решение:

1. $$\frac{b + 3a}{18a^2b} + \frac{a - 4b}{24ab^2} = \frac{4b(b + 3a) + 3a(a - 4b)}{72a^2b^2} = \frac{4b^2 + 12ab + 3a^2 - 12ab}{72a^2b^2} = \frac{4b^2 + 3a^2}{72a^2b^2}$$
2. $$\frac{a - 5}{5a + 25} + \frac{3a + 5}{a^2 + 5a} = \frac{a - 5}{5(a + 5)} + \frac{3a + 5}{a(a + 5)} = \frac{a(a - 5) + 5(3a + 5)}{5a(a + 5)} = \frac{a^2 - 5a + 15a + 25}{5a(a + 5)} = \frac{a^2 + 10a + 25}{5a(a + 5)} = \frac{(a + 5)^2}{5a(a + 5)} = \frac{a + 5}{5a}$$
3. $$\frac{m - 8}{5m} : \frac{m^2 - 64}{15m^2} = \frac{m - 8}{5m} \cdot \frac{15m^2}{m^2 - 64} = \frac{m - 8}{5m} \cdot \frac{15m^2}{(m - 8)(m + 8)} = \frac{15m^2}{5m(m + 8)} = \frac{3m}{m + 8}$$
4. $$(\frac{c - 2}{c + 2} - \frac{c}{c - 2}) \cdot \frac{c + 2}{2 - 3c} = (\frac{(c - 2)^2 - c(c + 2)}{(c + 2)(c - 2)}) \cdot \frac{c + 2}{2 - 3c} = (\frac{c^2 - 4c + 4 - c^2 - 2c}{(c + 2)(c - 2)}) \cdot \frac{c + 2}{2 - 3c} = (\frac{-6c + 4}{(c + 2)(c - 2)}) \cdot \frac{c + 2}{2 - 3c} = \frac{2(-3c + 2)}{(c + 2)(c - 2)} \cdot \frac{c + 2}{2 - 3c} = \frac{2}{c - 2}$$