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2. $\frac{a-\frac{4a-4}{a}}{ \frac{2}{a}-1}= \frac{\frac{a^2-(4a-4)}{a}}{ \frac{2-a}{a}}= \frac{(a^2-4a+4)a}{a(2-a)}=\frac{(a-2)^2}{2-a}=\frac{(2-a)^2}{2-a}=2-a; \\ a\in Q.$

3. $(\frac{1}{2-4b}+\frac{b+1}{8b^3-1}\cdot\frac{4b^2+2b+1}{1+2b}):\frac{1}{4b-2}= (-\frac{1}{4b-2}+\\ \frac{b+1}{(2b-1)(4b^2+2b+1)}\cdot\frac{4b^2+2b+1}{1+2b})\cdot(4b-2)= -\frac{4b-2}{4b-2}+\frac{(b+1)(4b-2)}{(2b-1)(1+2b)}= \\ = -1+\frac{2(b+1)(2b-1)}{(2b-1)(1+2b)}= -1+\frac{2(b+1)}{1+2b}=\frac{-1-2b+2b+2}{1+2b}=\frac{1}{1+2b}.$

4. $(\frac{2}{(1-x)^2} + \frac{1}{x^2-1})\cdot(x-1)^2- \frac{3x}{x+1} = (\frac{2(x-1)^2}{(x-1)^2} + \frac{(x-1)^2}{(x+1)(x-1)})- \frac{3x}{x+1} = \\ =2 + \frac{x-1}{x+1}- \frac{3x}{x+1} = \frac{2(x+1)+x-1-3x}{x+1} = \frac{1}{x+1}; \\ \frac{1}{-1,5+1}=\frac{1}{-0,5}=-2$

5. $(\frac{2}{(n+1)!} - \frac{3}{n!}):(\frac{5}{n!} - \frac{4n}{(n+1)!}) = (\frac{2}{n!\cdot(n+1)} - \frac{3}{n!}):(\frac{5}{n!} - \frac{4n}{n!(n+1)}) = \\ = \frac{2-3(n+1)}{n!\cdot(n+1)}:\frac{5(n+1)-4n}{n!\cdot(n+1)} = \frac{-3n-1}{(n+1)!}\cdot\frac{(n+1)!}{n+5} = \frac{-3n-1}{n+5} = \frac{-3(n+5)+14}{n+5} = \\ =\frac{-3(n+5)}{n+5}+\frac{14}{n+5} = -3+\frac{14}{n+5}; \\ n+5=14, \\ n=9.$