1. Given real numbers $a$, $b$, $x$, $y$ satisfy $x y=2008^{2008}$,
$$
\frac{1}{1+2008^{a} x}+\frac{1}{1+2008^{b-2009} y}=1 \text {. }
$$

Then the value of $2008^{a+b}$ is
Ni, 1.2008.
Given the equation, after removing the denominator, we get
$$
\begin{array}{l}
1+2008^{b-2009} y+1+2008^{a} x \\
=\left(1+2008^{a} x\right)\left(1+2008^{b-2009} y\right) .
\end{array}
$$

Simplifying, we get
$$
\begin{array}{l}
2008^{a+b-2009} x y=1 \\
\Rightarrow 2008^{a+b-2009} \times 2008^{2008}=1 \\
\Rightarrow 2008^{a+b-1}=1 .
\end{array}
$$

Therefore, $2008^{a+b}=2008$.