Помогите пожалуйста номер 2, 2 задание

Решите задачу:

$1)\quad \frac{y}{x}=\frac{1}{2}\\\\\frac{x^3+x^2y+xy^2+y^3}{x^3-y^3}=\left [\, \frac{:x^3}{:x^3}\, \right ]=\frac{1+\frac{y}{x}+(\frac{y}{x})^2+(\frac{y}{x})3}{1-(\frac{y}{x})^3}=\frac{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}{1-\frac{1}{8}}=\\\\=\frac{15/8}{7/8}=\frac{15}{7}=2\frac{1}{7}\\\\2)\quad b-\frac{7}{b}=3\\\\3^2=(b-\frac{7}{b})^2=b^2-2\cdot b\cdot \frac{7}{b}+\frac{49}{b^2}=b^2+\frac{49}{b^2}-14\; \; \to \\\\b^2+\frac{49}{b^2}=9+14=23$

$3)\quad b=-0,7\\\\(4+4b+b^2)(\frac{1}{b-2}-\frac{4}{b^2-4}-\frac{1}{2b+4})=(b+2)^2\cdot \frac{2(b+2)-2-(b-2)}{2(b-2)(b+2)}=\\\\=(b+2)\cdot \frac{2b+4-2-b+2}{2(b-2)}=\frac{(b+2)(b+4)}{2(b-2)}$