\(k)\left(-2\right)^3\cdot\dfrac{-1}{24}+\left(\dfrac{4}{5}-1,2\right):\dfrac{2}{15}\\ =-8\cdot\dfrac{-1}{24}+\left(\dfrac{4}{5}-\dfrac{6}{5}\right)\cdot\dfrac{15}{2}\\ =\dfrac{\left(-8\right)\cdot\left(-1\right)}{24}+\dfrac{-2}{5}\cdot\dfrac{15}{2}\\ =\dfrac{1}{3}+\left(-3\right)\\ =\dfrac{1}{3}-\dfrac{9}{3}\\ =-\dfrac{8}{3}\)
\(l)25\%-1\dfrac{1}{2}-\left(-\dfrac{1}{2}\right)^2+0,25:\dfrac{1}{12}\\ =\dfrac{1}{4}-\dfrac{3}{2}-\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{12}{1}\\ =\dfrac{1}{4}-\dfrac{6}{4}-\dfrac{1}{4}+3\\ =-\dfrac{3}{2}+\dfrac{6}{2}\\ =\dfrac{3}{2}\)
\(m)\left(\dfrac{-2}{5}\right)^2+\dfrac{1}{2}\cdot\left(4,5-2\right)-50\%\\ =\dfrac{4}{25}+\dfrac{1}{2}\cdot\left(\dfrac{9}{2}-\dfrac{4}{2}\right)-\dfrac{1}{2}\\ =\dfrac{4}{25}+\dfrac{5}{4}-\dfrac{1}{2}\\ =\dfrac{16}{100}+\dfrac{125}{100}-\dfrac{50}{100}\\ =\dfrac{91}{100}.\)