Помогите с номером 1033.

Решите задачу:

$1)\;8\frac7{10}-\left(7\frac45-a\right)=8\frac7{10}-7\frac8{10}+a=\frac9{10}+a=\frac9{10}-6\frac12=\\=\frac9{10}+6\frac5{10}=7\frac4{10}=7\frac25\\\\2)\;14\frac16-\left(8\frac9{10}-b\right)=14\frac5{30}-8\frac{27}{30}+b=5\frac{8}{30}+b=5\frac{4}{15}+5\frac23=\\=5\frac{4}{15}+5\frac{10}{15}=11\frac{14}{15}\\\\3)\;a-\left(9\frac3{50}-5\frac7{25}\right)=a-\left(9\frac3{50}-5\frac{14}{50}\right)=a-3\frac{39}{50}=7\frac12-3\frac{39}{50}=\\=7\frac{25}{50}-3\frac{39}{50}=3\frac{36}{50}=3\frac{18}{25}$

$4)\;\left(20\frac15-b\right)-6\frac13=20\frac3{15}-b-6\frac5{15}=13\frac{13}{15}-b=13\frac{13}{15}-11\frac2{15}=2\frac{11}{15}$