$1.39 \frac{\sqrt{2 b+2 \sqrt{b^{2}-4}}}{\sqrt{b^{2}-4}+b+2}$.
1.39 It is easy to guess that the expression under the square root in the numerator is a complete square of the sum: $(\sqrt{b+2}+\sqrt{b-2})^{2}$. Then we get

$$
\frac{\sqrt{b+2}+\sqrt{b-2}}{\sqrt{b^{2}-4}+b+2}=\frac{\sqrt{b+2}+\sqrt{b-2}}{\sqrt{b+2}(\sqrt{b+2}+\sqrt{b-2})}=\frac{1}{\sqrt{b+2}}
$$

Answer: $\frac{1}{\sqrt{b+2}}$.