Решение:

1) a) $$\begin{aligned} (\frac{2a}{b^2} - \frac{1}{2a}) : (\frac{1}{b} + \frac{1}{2a}) &= \frac{4a^2 - b^2}{2ab^2} : \frac{2a + b}{2ab} = \frac{(2a - b)(2a + b)}{2ab^2} \cdot \frac{2ab}{2a + b} = \frac{2a - b}{b} \end{aligned}$$

Ответ: $$\frac{2a - b}{b}$$

в) $$\begin{aligned} \frac{y-3}{y+3} \cdot (y + \frac{y^2}{3-y}) &= \frac{y-3}{y+3} \cdot (y - \frac{y^2}{y-3}) = \frac{y-3}{y+3} \cdot \frac{y(y-3) - y^2}{y-3} = \frac{y-3}{y+3} \cdot \frac{y^2 - 3y - y^2}{y-3} = \frac{-3y}{y+3} \end{aligned}$$

Ответ: $$\frac{-3y}{y+3}$$

д) $$\begin{aligned} \frac{6x+y}{3x} - \frac{5y^2}{x^2} \cdot \frac{x}{15y} &= \frac{6x+y}{3x} - \frac{5y^2x}{15x^2y} = \frac{6x+y}{3x} - \frac{y}{3x} = \frac{6x + y - y}{3x} = \frac{6x}{3x} = 2 \end{aligned}$$

Ответ: 2

2) a) $$\begin{aligned} \frac{a^2-x^2}{b^2-16} \cdot \frac{b+4}{a-x} + \frac{x}{4-b} &= \frac{(a-x)(a+x)}{(b-4)(b+4)} \cdot \frac{b+4}{a-x} - \frac{x}{b-4} = \frac{a+x}{b-4} - \frac{x}{b-4} = \frac{a+x-x}{b-4} = \frac{a}{b-4} \end{aligned}$$

Ответ: $$\frac{a}{b-4}$$

в) $$\begin{aligned} (\frac{2a^2-a-2}{a^2-a+1} - 2) : (\frac{1}{a+1} - \frac{a-1}{a^2-a+1}) &= (\frac{2a^2-a-2 - 2(a^2-a+1)}{a^2-a+1}) : (\frac{a^2-a+1 - (a-1)(a+1)}{(a+1)(a^2-a+1)}) = \frac{2a^2-a-2 - 2a^2+2a-2}{a^2-a+1} : \frac{a^2-a+1 - a^2+1}{(a+1)(a^2-a+1)} = \frac{a-4}{a^2-a+1} : \frac{-a+2}{(a+1)(a^2-a+1)} = \frac{a-4}{a^2-a+1} \cdot \frac{(a+1)(a^2-a+1)}{-a+2} = \frac{(a-4)(a+1)}{-a+2} = -\frac{(a-4)(a+1)}{a-2} \end{aligned}$$

Ответ: $$\frac{(a-4)(a+1)}{-(a-2)}$$