[tex]\Big(1\dfrac{2}{3}+0,5\Big)\cdot\Big(5\dfrac{2}{5}-3,9\Big)=\Big(\dfrac{5}{3}+\dfrac{1}{2}\Big)\cdot(5,4-3,9)=\\\\\\=\Big(\dfrac{10}{6}+\dfrac{3}{6}\Big)\cdot1,5=\dfrac{13}{6}\cdot\dfrac{3}{2}=\dfrac{13}{4}=\boxed{3\dfrac{1}{4}}[/tex]
[tex]\Big(1,5+\dfrac{2}{3}\Big)\cdot\Big(6\dfrac{2}{5}-4,9\Big)=\Big(\dfrac{3}{2}+\dfrac{2}{3}\Big)\cdot(6,4-4,9)=\\\\\\=\Big(\dfrac{9}{6}+\dfrac{4}{6}\Big)\cdot1,5=\dfrac{13}{6}\cdot\dfrac{3}{2}=\dfrac{13}{4}=\boxed{3\dfrac{1}{4}}[/tex]