5) $$\frac{\left(\frac{7}{15}+\frac{14}{45}+\frac{2}{9}\right)\cdot 10\frac{1}{3}-1\frac{1}{11}\left(2\frac{2}{3}-1.75\right)}{\left(\frac{3}{7}-0.25\right):\frac{3}{28}-1}$$;

$$\frac{\left(\frac{7}{15}+\frac{14}{45}+\frac{2}{9}\right)\cdot 10\frac{1}{3}-1\frac{1}{11}\left(2\frac{2}{3}-1.75\right)}{\left(\frac{3}{7}-0.25\right):\frac{3}{28}-1}=\frac{\left(\frac{21+14+10}{45}\right)\cdot \frac{31}{3}-\frac{12}{11}\left(\frac{8}{3}-1.75\right)}{\left(\frac{3}{7}-\frac{1}{4}\right):\frac{3}{28}-1}=\frac{\frac{45}{45}\cdot \frac{31}{3}-\frac{12}{11}\left(\frac{32-21}{12}\right)}{\left(\frac{12-7}{28}\right):\frac{3}{28}-1}=\frac{\frac{31}{3}-\frac{12}{11}\cdot \frac{11}{12}}{\frac{5}{28}:\frac{3}{28}-1}=\frac{\frac{31}{3}-1}{\frac{5}{3}-1}=\frac{\frac{28}{3}}{\frac{2}{3}}=14$$