$a)\frac{6\frac{3}{4}\cdot \left(2\frac{1}{3}-1\frac{5}{6}\right)-\frac{5}{8}}{3\frac{3}{4}\cdot \frac{11}{15}-\frac{1}{2}}=\frac{6\frac{3}{4}\cdot \left(2\frac{2}{6}-1\frac{5}{6}\right)-\frac{5}{8}}{\frac{{\sout{15}}^1}{4}\cdot \frac{11}{{\sout{15}}^1}-\frac{1}{2}}=\frac{6\frac{3}{4}\cdot \left(1\frac{8}{6}-1\frac{5}{6}\right)-\frac{5}{8}}{\frac{11}{4}-\frac{2}{4}}=$

$=\frac{6\frac{3}{4}\cdot \frac{3}{6}-\frac{5}{8}}{\frac{9}{4}}=\left(\frac{27}{4}\cdot \frac{1}{2}-\frac{5}{8}\right):\frac{9}{4}=$

$=\left(\frac{27}{8}-\frac{5}{8}\right)\cdot \frac{4}{9}=\frac{22}{8}\cdot \frac{4}{9}=\frac{11}{{\sout{4}}^1}\cdot \frac{{\sout{4}}^1}{9}=\frac{11}{9}=1\frac{2}{9}$

 

 

$b)\frac{0,824\cdot \left(3,57+11,43\right)}{4,72\cdot 0,15+0,492}=\frac{0,824\cdot 15}{0,708+0,492}=\frac{12,36}{1,2}=\frac{123,6}{12}=10,3$

 

 

             

 

$c)2,8+2,8\cdot 3\frac{5}{7}+2,8\cdot 3\frac{5}{7}\cdot \frac{35}{52}+2,8\cdot 3\frac{5}{7}\cdot \frac{35}{52}\cdot 0,55=$

$=2,8\cdot \left(1+3\frac{5}{7}+3\frac{5}{7}\cdot \frac{35}{52}+3\frac{5}{7}\cdot \frac{35}{52}\cdot 0,55\right)=$

$=2,8\cdot \left(1+3\frac{5}{7}+\frac{{\sout{26}}^1}{7}\cdot \frac{35}{{\sout{52}}^2}+\frac{{\sout{26}}^1}{7}\cdot \frac{35}{{\sout{52}}^2}\cdot \frac{55}{100}\right)=$

$=2,8\cdot \left(1+3\frac{5}{7}+\frac{35}{14}+\frac{35}{14}\cdot \frac{11}{20}\right)=$

$=2,8\cdot \left(4\frac{5}{7}+\frac{5}{2}+\frac{{\sout{5}}^1}{2}\cdot \frac{11}{{\sout{20}}^4}\right)=$

$=2,8\cdot \left(4\frac{5}{7}+2\frac{1}{2}+\frac{11}{8}\right)=2,8\cdot \left(4\frac{5}{7}+2\frac{4}{8}+1\frac{3}{8}\right)=$

$=\frac{28}{10}\cdot \left(4\frac{5}{7}+3\frac{7}{8}\right)=\frac{28}{10}\cdot \left(4\frac{40}{56}+3\frac{49}{56}\right)=$

$=\frac{28}{10}\cdot 7\frac{89}{56}=\frac{{\sout{28}}^1}{10}\cdot \frac{7\cdot 56+89}{{\sout{56}}^2}=$

$=\frac{392+89}{20}=\frac{481}{20}=\frac{240,5}{10}=24,05$