2) 8+61-4,3;

2) $$8+\frac{7}{9}+6\frac{2}{15}-4.3=8+\frac{7}{9}+6\frac{2}{15}-\frac{43}{10}=14+\frac{7}{9}+\frac{2}{15}-\frac{3}{10}=14+\frac{7 \cdot 10}{9 \cdot 10}+\frac{2 \cdot 6}{15 \cdot 6}-\frac{43 \cdot 9}{10 \cdot 9}=14+\frac{70}{90}+\frac{12}{90}-\frac{387}{90}=14+\frac{70+12-387}{90}=14-\frac{305}{90}=14-\frac{61}{18}=\frac{14 \cdot 18}{18}-\frac{61}{18}=\frac{252-61}{18}=\frac{191}{18}=10\frac{11}{18}$$

Ответ: $$10\frac{11}{18}$$