a)

$\begin{array}{l}0,3-\frac{4}{9}:\frac{4}{3}\cdot \frac{6}{5}+1\\ =\frac{3}{10}-\frac{4}{9}.\frac{3}{4}.\frac{6}{5}+1\\ =\frac{3}{10}-\frac{2}{5}+1\\ =\frac{3}{10}-\frac{4}{10}+\frac{10}{10}\\ =\frac{9}{10}\end{array}$

b)

$\begin{array}{l}{\left(\frac{-1}{3}\right)}^2-\frac{3}{8}:(0,5{)}^3-\frac{5}{2}\cdot (-4)\\ =\frac{1}{9}-\frac{3}{8}:\frac{1}{8}-\frac{5}{2}.\left(-4\right)\\ =\frac{1}{9}-3+10\\ =\frac{1}{9}-\frac{27}{9}+\frac{90}{9}\\ =\frac{64}{9}\end{array}$

c)

$\begin{array}{l}1+2:\left(\frac{2}{3}-\frac{1}{6}\right)\cdot (-2,25)\\ =1+2:\left(\frac{4}{6}-\frac{1}{6}\right)\cdot \left(-\frac{9}{4}\right)\\ =1+2:\frac{1}{2}.\left(-\frac{9}{4}\right)\\ =1+2.\frac{2}{1}.\left(-\frac{9}{4}\right)\\ =1+\left(-9\right)=-8\end{array}$$\begin{array}{l}1+2:\left(\frac{2}{3}-\frac{1}{6}\right)\cdot (-2,25)\\ =1+2:\left(\frac{4}{6}-\frac{1}{6}\right)\cdot \left(-\frac{9}{4}\right)\\ =1+2:\frac{1}{2}.\left(-\frac{9}{4}\right)\\ =1+2.\frac{2}{1}.\left(-\frac{9}{4}\right)\\ =1+\left(-9\right)=-8\end{array}$

d)

$\begin{array}{l}\left[\left(\frac{1}{4}-0,5\right)\cdot 2+\frac{8}{3}\right]:2\\ =\left[\left(\frac{1}{4}-\frac{2}{4}\right)\cdot 2+\frac{8}{3}\right].\frac{1}{2}\\ =\left(\frac{-1}{4}.2+\frac{8}{3}\right).\frac{1}{2}\\ =\left(\frac{-1}{2}+\frac{8}{3}\right).\frac{1}{2}\\ =\frac{13}{6}.\frac{1}{2}=\frac{13}{12}\end{array}$