$$
\begin{array}{l}
\text { 1. Let } A=\left\{x \in \mathbf{Z} \left\lvert\, \frac{x^{2}}{2009}+\frac{y^{2}}{2008}=1\right.\right\}, \\
B=\{x \mid x=2 n+1, n \in \mathbf{Z}\},
\end{array}
$$

Set $M$ is a subset of $A$, but not a subset of $B$. Then the number of all such sets $M$ is $\qquad$
$$
\begin{array}{l}
-1.2^{41}\left(2^{45}-1\right) . \\
\text { From } \frac{x^{2}}{2009}+\frac{y^{2}}{2008}=1(x \in \mathbf{Z}) \text { we get }|x| \leqslant 44 \text {. }
\end{array}
$$

Therefore, $A=\{-44,-43, \cdots,-1,0,1, \cdots, 44\}$.
Obviously, $|A|=89$, and the subsets of $A$ total $2^{89}$, among which the subsets of $B$ are
$$
\{-43,-41, \cdots,-1,1, \cdots, 43\}
$$

with $2^{44}$ subsets.
Thus, the number of sets $M$ that meet the conditions is
$$
2^{89}-2^{44}=2^{44}\left(2^{45}-1\right) \text {. }
$$