14.
[tex]a)\\\\(2\frac{2}{3}\cdot\frac{7}{8}-1\frac{1}{2}):\frac{4}{9} = (\frac{8}{3}\cdot\frac{7}{8}-\frac{3}{2})\cdot\frac{9}{4}=(\frac{7}{3}-\frac{3}{2})\cdot\frac{9}{4} = (\frac{14}{6}-\frac{9}{6})\cdot\frac{9}{4} =\frac{5}{6}\cdot\frac{9}{4} = \frac{5}{2}\cdot\frac{3}{4} =\frac{15}{8} =1\frac{7}{8}[/tex]
[tex]b)\\\\5\frac{1}{6}:[12\cdot(1\frac{5}{6}+\frac{3}{4})] = \frac{31}{6}:[12\cdot(\frac{11}{6}+\frac{3}{4})] = \frac{31}{6}:[12\cdot(\frac{22}{12}+\frac{9}{12})] = \frac{31}{6}:(12\cdot\frac{31}{12})=\\\\=\frac{31}{6}:31=\frac{31}{6}\cdot\frac{1}{31} = \frac{1}{6}[/tex]