3. 82 .
$$\begin{aligned} \text { Given } & \frac{(1-a)^{2}}{b}-\frac{(1+b)^{2}}{a}=4 \\ \Rightarrow & a(1-a)^{2}-b(1+b)^{2}=4 a b \\ \Rightarrow & a-2 a^{2}+a^{3}-b-2 b^{2}-b^{3}=4 a b \\ \Rightarrow & (a-b)-2\left(a^{2}+b^{2}\right)+\left(a^{3}-b^{3}\right)=4 a b \\ \Rightarrow & (a-b)-2\left[(a-b)^{2}+2 a b\right]+ \\ & (a-b)\left[(a-b)^{2}+3 a b\right]=4 a b . \end{aligned}$$

Substituting $a-b=2$ into the equation, we get $a b=1$.
$$\begin{array}{l} \text { Then } a^{2}+b^{2}=(a-b)^{2}+2 a b=6, \\ a^{3}-b^{3}=(a-b)\left[(a-b)^{2}+3 a b\right]=14. \\ \text { Therefore, } a^{5}-b^{5}=\left(a^{2}+b^{2}\right)\left(a^{3}-b^{3}\right)-a^{2} b^{2}(a-b) \\ =6 \times 14-1 \times 2=82. \end{array}$$