Вычислите: 1) \(\frac{9}{52}\cdot 4\frac{1}{3}:1\frac{1}{3}+\left(3\frac{2}{3}+1\frac{4}{5}\right)\cdot\left(\frac{60}{97}+\frac{5}{36}\cdot\frac{14}{15}\right)\); 2) \(\left(4\frac{1}{3}+2\frac{3}{15}\right)\cdot\left(4\frac{1}{5}-1\frac{8}{25}\right)\); 3) \(\left(8\frac{11}{24}-4\frac{1}{12}\right)\cdot\left(3\frac{1}{4}+1\frac{1}{28}\right)\).

\(\frac{9}{52}\cdot 4\frac{1}{3}:1\frac{1}{3}+\left(3\frac{2}{3}+1\frac{4}{5}\right)\cdot\left(\frac{60}{97}+\frac{5}{36}\cdot\frac{14}{15}\right)\)

Преобразуем смешанные числа:

\(4\frac{1}{3}=\frac{13}{3}\), \(1\frac{1}{3}=\frac{4}{3}\), \(3\frac{2}{3}=\frac{11}{3}\), \(1\frac{4}{5}=\frac{9}{5}\).

\(\frac{9}{52}\cdot\frac{13}{3}:\frac{4}{3}=\frac{9}{52}\cdot\frac{13}{4}=\frac{9}{16}\).

\(\frac{11}{3}+\frac{9}{5}=\frac{82}{15}\).

\(\frac{5}{36}\cdot\frac{14}{15}=\frac{7}{54}\), поэтому

\(\frac{60}{97}+\frac{7}{54}=\frac{3919}{5238}\).

Тогда

\(\frac{9}{16}+\frac{82}{15}\cdot\frac{3919}{5238}=\frac{9}{16}+\frac{160679}{39285}=\frac{2924429}{628560}\).

Ответ: \(\frac{2924429}{628560}=4\frac{403469}{628560}\).

\(\left(4\frac{1}{3}+2\frac{3}{15}\right)\cdot\left(4\frac{1}{5}-1\frac{8}{25}\right)\)

\(4\frac{1}{3}+2\frac{3}{15}=\frac{13}{3}+\frac{11}{5}=\frac{98}{15}\).

\(4\frac{1}{5}-1\frac{8}{25}=\frac{21}{5}-\frac{33}{25}=\frac{72}{25}\).

\(\frac{98}{15}\cdot\frac{72}{25}=\frac{7056}{375}=\frac{2352}{125}=18\frac{102}{125}\).

Ответ: \(18\frac{102}{125}\).

\(\left(8\frac{11}{24}-4\frac{1}{12}\right)\cdot\left(3\frac{1}{4}+1\frac{1}{28}\right)\)

\(8\frac{11}{24}-4\frac{1}{12}=\frac{203}{24}-\frac{49}{12}=\frac{203}{24}-\frac{98}{24}=\frac{105}{24}=\frac{35}{8}\).

\(3\frac{1}{4}+1\frac{1}{28}=\frac{13}{4}+\frac{29}{28}=\frac{91}{28}+\frac{29}{28}=\frac{30}{7}\).

\(\frac{35}{8}\cdot\frac{30}{7}=\frac{150}{8}=\frac{75}{4}=18\frac{3}{4}\).

Ответ: \(18\frac{3}{4}\).