д) \(\frac{16}{27}\) \(\cdot\) (\(\frac{25}{72}\) + \(\frac{11}{72}\)) = ...

1. Вынесем общий множитель \(\frac{16}{27}\) за скобки: \[\frac{16}{27} \cdot \frac{25}{72} + \frac{16}{27} \cdot \frac{11}{72} = \frac{16}{27} \cdot \left(\frac{25}{72} - \frac{11}{72}\right).\]
2. Упростим выражение в скобках: \[\frac{16}{27} \cdot \left(\frac{25}{72} - \frac{11}{72}\right) = \frac{16}{27} \cdot \left(\frac{25 - 11}{72}\right) = \frac{16}{27} \cdot \frac{14}{72}.\]
3. Сократим дроби: \[\frac{16}{27} \cdot \frac{14}{72} = \frac{8}{27} \cdot \frac{14}{36} = \frac{4}{27} \cdot \frac{14}{18} = \frac{4}{27} \cdot \frac{7}{9}.\]
4. Выполним умножение: \[\frac{4}{27} \cdot \frac{7}{9} = \frac{4 \cdot 7}{27 \cdot 9} = \frac{28}{243}.\]

Ответ: \(\frac{28}{243}\)