$\frac{\frac{2a}{1 - a}}{1 - (\frac{1 - a}{2a})^{-1}} = \frac{\frac{2a}{1 - a}}{1 - \frac{2a}{1 - a}} = \frac{\frac{2a}{1 - a}}{\frac{1 - a - 2a}{1 - a}} = \frac{\frac{2a}{1 - a}}{\frac{1 - 3a}{1 - a}} = \frac{2a}{1 - a} : \frac{1 - 3a}{1 - a} = \frac{2a}{1 - a} * \frac{1 - a}{1 - 3a} = \frac{2a}{1 - 3a}$

$\frac{\frac{2a}{2 - a}}{2 - (\frac{2 - a}{2a})^{-1}} = \frac{\frac{2a}{2 - a}}{2 - \frac{2a}{2 - a}} = \frac{\frac{2a}{2 - a}}{\frac{4 - 2a - 2a}{2 - a}} = \frac{\frac{2a}{2 - a}}{\frac{4 - 4a}{2 - a}} = \frac{2a}{2 - a} : \frac{4 - 4a}{2 - a} = \frac{2a}{2 - a} * \frac{2 - a}{4(1 - a)} = \frac{a}{2(2 - a)}$

$(\frac{1}{x} - \frac{1}{x + 3})^{-1} + (\frac{3}{x + 3} - \frac{3}{x})^{-1} = (\frac{x + 3 - x}{x(x + 3)})^{-1} + (\frac{3x - 3x - 9}{x(x + 3)})^{-1} = (\frac{3}{x(x + 3)})^{-1} - (\frac{-9}{x(x + 3)})^{-1} = \frac{x(x + 3)}{3} - \frac{x(x + 3)}{9} = \frac{3x^2 + 9x - x^2 - 3x}{9} = \frac{2x^2 + 6x}{9} = \frac{2x(x + 3)}{9}$

$(\frac{1}{x} - \frac{1}{x - 1})^{-1} + (\frac{4}{x - 1} - \frac{4}{x})^{-1} = (\frac{x - 1 - x}{x(x - 1)})^{-1} + (\frac{4x - 4x + 4}{x(x - 1)})^{-1} = (\frac{-1}{x(x - 1)})^{-1} + (\frac{4}{x(x - 1)})^{-1} = -\frac{x(x - 1)}{1} + \frac{x(x - 1)}{4} = \frac{-4x^2 + 4x + x^2 - x}{4} = \frac{-3x^2 + 3x}{4} = \frac{-3x(x - 1)}{4}$