1, \(\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}=\left(\dfrac{2}{3}-\dfrac{3}{2}\right)\cdot\dfrac{3}{4}+\dfrac{1}{2}=-\dfrac{5}{6}\cdot\dfrac{3}{4}+\dfrac{1}{2}=-\dfrac{5}{8}+\dfrac{1}{2}=-\dfrac{1}{8}\)
2, \(\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}=\left(-1\right)+1-\dfrac{3}{7}=0-\dfrac{3}{7}=-\dfrac{3}{7}\)
\(\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}=\dfrac{-5}{6}x\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{-5}{8}+\dfrac{1}{2}=\dfrac{-1}{8}\)
\(\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}=-1+1-\dfrac{3}{7}=0-\dfrac{3}{7}=\dfrac{-3}{7}\)