$\text{a)}6:\frac{1}{3}-0,8:\frac{1,5}{\frac{3}{2}\cdot 0,4\cdot \frac{50}{1:\frac{1}{2}}}+\frac{1}{4}+\frac{1+\frac{1}{2}\cdot \frac{1}{0,25}}{6-\frac{46}{1+2,2\cdot 10}}=6\cdot \frac{3}{1}-0,8:\frac{1,5}{\frac{1,2}{2}\cdot \frac{50}{1\cdot \frac{2}{1}}}+\frac{1}{4}+\frac{1+\frac{1}{0,5}}{6-\frac{46}{1+22}}=18-0,8:\frac{1,5}{0,6\cdot \frac{50}{2}}+\frac{1}{4}+\frac{1+\frac{10}{5}}{6-\frac{46}{23}}=18-0,8:\frac{1,5}{0,6\cdot 25}+\frac{1}{4}+\frac{1+2}{6-2}=18-0,8:\frac{1,5}{15}+\frac{1}{4}+\frac{3}{4}=18-0,8:\frac{15}{150}+1=18-0,8:\frac{1}{10}+1=18-0,8\cdot \frac{10}{1}+1=18-8+1=11$  

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$\text{b}\frac{\left[{\left(1\frac{1}{3}\right)}^{-2}-\frac{10}{2^4}\right]:{\left(-2\right)}^{-5}}{{\left[{\left(3\frac{1}{3}\right)}^{-1}\cdot 3^0-{1,25}^{-1}\cdot {\left(\frac{3}{2}\right)}^2\right]}^{-1}}=\frac{\left[{\left(\frac{4}{3}\right)}^{-2}-\frac{10}{16}\right]:\frac{1}{{\left(-2\right)}^5}}{{\left[{\left(\frac{10}{3}\right)}^{-1}\cdot 1-{\left(1\frac{1}{4}\right)}^{-1}\cdot \frac{9}{4}\right]}^{-1}}=\frac{\left[{\left(\frac{3}{4}\right)}^2-\frac{10}{16}\right]:\frac{1}{-32}}{{\left[\frac{3}{10}-{\left(\frac{5}{4}\right)}^{-1}\cdot \frac{9}{4}\right]}^{-1}}=\frac{\left(\frac{9}{16}-\frac{10}{16}\right)\cdot \left(-\frac{32}{1}\right)}{{\left(\frac{3}{10}-\frac{{\sout{4}}^1}{5}\cdot \frac{9}{{\sout{4}}^1}\right)}^{-1}}=\frac{-\frac{1}{16}\cdot \left(-\frac{32}{1}\right)}{{\left(\frac{3}{10}-\frac{9}{5}\right)}^{-1}}=\frac{\frac{32}{16}}{{\left(\frac{3}{10}-\frac{18}{10}\right)}^{-1}}=\frac{2}{{\left(-\frac{15}{10}\right)}^{-1}}=\frac{2}{-\frac{10}{15}}=2:\left(-\frac{10}{15}\right)=2\cdot \left(-\frac{15}{10}\right)=2\cdot \left(-\frac{3}{2}\right)=-3$