Для пункта (a): \(\frac{1}{4} + \frac{5}{14} = \frac{7}{28} + \frac{10}{28} = \frac{17}{28}\), \(\frac{3}{5} + \frac{6}{10} = \frac{6}{10} + \frac{6}{10} = \frac{12}{10}\). Деление: \(\frac{17}{28} \div \frac{12}{10} = \frac{17}{28} \cdot \frac{10}{12} = \frac{170}{336} = \frac{85}{168}\). Для пункта (b): \(\frac{1}{7} \div (\frac{1}{3} + \frac{8}{15}) = \frac{1}{7} \div \frac{13}{15} = \frac{15}{91}\), \(\frac{5}{12} \div \frac{1}{10} = \frac{50}{12} = \frac{25}{6}\). Результат: \(\frac{15}{91} - \frac{25}{6} = -\frac{1961}{546}\).