$\text{a)}4,67-\left(3\frac{5}{6}-\frac{1}{3}\right)=4,67-\left(3\frac{5}{6}-\frac{2}{6}\right)=4,67-3\frac{3}{6}=4,67-3\frac{1}{2}=4,67-3,5=1,17$  

$\text{b)}2,8+1\frac{1}{3}\cdot 0,9=2,8+\frac{4}{{\sout{3}}^1}\cdot \frac{{\sout{9}}^3}{10}=2,8+\frac{4}{1}\cdot \frac{3}{10}=2,8+\frac{12}{10}=2,8+1,2=4$  

$\text{c)}3,6:\frac{3}{5}-1\frac{1}{5}=\frac{{\sout{36}}^{12}}{10}\cdot \frac{5}{{\sout{3}}^1}-1\frac{1}{5}=\frac{12}{{\sout{10}}^2}\cdot \frac{{\sout{5}}^1}{1}-1\frac{1}{5}=\frac{12}{2}\cdot \frac{1}{1}-1\frac{1}{5}=6-1\frac{1}{5}=4\frac{4}{5}$  

 

$\text{d)}{0,2}^2\cdot \left(1,6-\frac{2}{5}\right)=0,04\cdot \left(1,6-\frac{4}{10}\right)=0,04\cdot \left(1,6-0,4\right)=0,04\cdot 1,2=0,048$    

 

$\text{e)}2,8\cdot 1\frac{1}{4}-\frac{2}{3}\cdot 0,75=\frac{{\sout{28}}^7}{10}\cdot \frac{5}{{\sout{4}}^1}-\frac{2}{{\sout{3}}^1}\cdot \frac{{\sout{3}}^1}{4}=\frac{7}{{\sout{10}}^2}\cdot \frac{{\sout{5}}^1}{1}-\frac{{\sout{2}}^1}{1}\cdot \frac{1}{{\sout{4}}^2}=\frac{7}{2}\cdot \frac{1}{1}-\frac{1}{1}\cdot \frac{1}{2}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3$  

 

$\text{f)}\left(4,5-1,1\right)\cdot 2\cdot \frac{3}{4}-1=3,4\cdot 2\cdot \frac{3}{4}-1=6,8\cdot \frac{3}{4}-1=\frac{{\sout{68}}^{17}}{10}\cdot \frac{3}{{\sout{4}}^1}-1=\frac{17}{10}\cdot \frac{3}{1}-1=\frac{51}{10}-1=5,1-1=4,1$