1. Simplify the expression: $\frac{8}{1+a^{8}}+\frac{4}{1+a^{4}}+\frac{2}{1+a^{2}}+\frac{1}{1+a}+\frac{1}{1-a}$ and find its value at $a=2^{-\frac{1}{16}}$.
Solution. Add the last two terms, then add the obtained sum to the third term from the end, and so on. As a result, we get $\frac{16}{1-a^{16}}$. Substituting $a$ with the value $2^{-\frac{1}{16}}$, we get 32.

Answer: 32