$a\text{)}2\frac{1}{8}\cdot \frac{16}{51}-7\frac{1}{4}\cdot \frac{10}{29}=\frac{17}{{\sout{8}}_1}\cdot \frac{{\sout{16}}^2}{51}-\frac{{\sout{29}}^1}{4}\cdot \frac{10}{{\sout{29}}_1}=\frac{34}{51}-\frac{10}{4}=\frac{2}{3}-\frac{5}{2}=\frac{4}{6}-\frac{15}{6}=-\frac{11}{6}$    

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$b\text{)}3\frac{3}{4}\cdot \frac{8}{25}+3\frac{2}{5}:\frac{34}{35}=\frac{15}{4}\cdot \frac{8}{25}+\frac{17}{5}:\frac{34}{35}=\frac{15}{{\sout{4}}_1}\cdot \frac{{\sout{8}}^2}{25}+\frac{{\sout{17}}^1}{5}\cdot \frac{35}{{\sout{34}}_2}=\frac{30}{25}+\frac{35}{10}=\frac{6}{5}+\frac{35}{10}=\frac{12}{10}+\frac{35}{10}=\frac{47}{10}=4,7$  

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$c\text{)}\frac{5}{12}:2\frac{7}{24}-\frac{7}{19}\cdot 2\frac{1}{28}=\frac{5}{12}:\frac{55}{24}-\frac{7}{19}\cdot \frac{57}{28}=\frac{5}{{\sout{12}}_1}\cdot \frac{{\sout{24}}^2}{55}-\frac{7}{{\sout{19}}_1}\cdot \frac{{\sout{57}}^3}{28}=\frac{10}{55}-\frac{21}{28}=\frac{2}{11}-\frac{3}{4}=\frac{8}{44}-\frac{33}{44}=-\frac{25}{44}$