88. 1) \(\left(1+\frac{1}{a}\right):\left(1-\frac{1}{a}\right)\); 2) \(\left(a+\frac{a}{b}\right)\left(a-\frac{a}{b}\right)\); 3) \(\left(\frac{b}{a}+\frac{a}{b}-2\right)\left(\frac{1}{b}-\frac{1}{a}\right)\); 4) \(\left(\frac{m}{n}+\frac{n}{m}+2\right)\left(1+\frac{m-n}{m+n}\right)\).

1) \(\left(1+\frac{1}{a}\right):\left(1-\frac{1}{a}\right)=\frac{a+1}{a}:\frac{a-1}{a}=\frac{a+1}{a-1}\).

2) \(\left(a+\frac{a}{b}\right)\left(a-\frac{a}{b}\right)=a^2-\frac{a^2}{b^2}=\frac{a^2(b^2-1)}{b^2}\).

3) \(\frac{b}{a}+\frac{a}{b}-2=\frac{(a-b)^2}{ab}\), а \(\frac{1}{b}-\frac{1}{a}=\frac{a-b}{ab}\). Поэтому \(\frac{(a-b)^3}{a^2b^2}\).

4) \(\frac{m}{n}+\frac{n}{m}+2=\frac{(m+n)^2}{mn}\), а \(1+\frac{m-n}{m+n}=\frac{2m}{m+n}\). Следовательно, \(\frac{(m+n)^2}{mn}\cdot\frac{2m}{m+n}=\frac{2(m+n)}{n}\).

Ответ: 1) \(\frac{a+1}{a-1}\); 2) \(\frac{a^2(b^2-1)}{b^2}\); 3) \(\frac{(a-b)^3}{a^2b^2}\); 4) \(\frac{2(m+n)}{n}\).