1. Let $x_{i} \in\{0,1\}(i=1,2, \cdots, n)$. If the function $f=f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ takes values only 0 or 1, then $f$ is called an $n$-ary Boolean function, and we denote
$$
D_{n}(f)=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=0\right\} .
$$
(1) Find the number of $n$-ary Boolean functions;
(2) Let $g$ be a 10-ary Boolean function, satisfying
$$
g\left(x_{1}, x_{2}, \cdots, x_{10}\right) \equiv 1+\sum_{i=1}^{10} \prod_{j=1}^{i} x_{j}(\bmod 2),
$$

Find the number of elements in the set $D_{10}(g)$, and find
$$
\sum_{\left(x_{1}, x_{2}, \cdots, x_{10}\right) \in D_{10}(g)}\left(x_{1}+x_{2}+\cdots+x_{10}\right) .
$$
1. (1) The total number of all possible values of $x_{1}, x_{2}, \cdots, x_{n}$ is $2^{n}$, and each corresponding function value can be either 0 or 1. Therefore, the number of all different $n$-ary Boolean functions is $2^{2^{n}}$.
(2) Let | $D_{10}(g)$ | denote the number of elements in the set $D_{10}(g)$. Below, “ * ” represents a value that can be 0 or 1. Then
$$
\begin{array}{l}
\text { when }\left(x_{1}, x_{2}, \cdots, x_{10}\right) \\
=(1,0, *, *, *, *, *, *, *, *)
\end{array}
$$

$g\left(x_{1}, x_{2}, \cdots, x_{10}\right)=0$, there are $2^{8}=256$ such cases;
$$
\begin{array}{l}
\text { when }\left(x_{1}, x_{2}, \cdots, x_{10}\right) \\
=(1,1,1,0, *, *, *, *, *, *)
\end{array}
$$

$g\left(x_{1}, x_{2}, \cdots, x_{10}\right)=0$, there are $2^{6}=64$ such cases;
$$
\begin{array}{l}
\text { when }\left(x_{1}, x_{2}, \cdots, x_{10}\right) \\
=(1,1,1,1,1,0, *, *, *, *)
\end{array}
$$

$g\left(x_{1}, x_{2}, \cdots, x_{10}\right)=0$, there are $2^{4}=16$ such cases;
$$
\begin{array}{l}
\text { when }\left(x_{1}, x_{2}, \cdots, x_{10}\right) \\
=(1,1,1,1,1,1,1,0, *, *)
\end{array}
$$

$g\left(x_{1}, x_{2}, \cdots, x_{10}\right)=0$, there are $2^{2}=4$ such cases;
$$
\begin{array}{l}
\text { when }\left(x_{1}, x_{2}, \cdots, x_{10}\right) \\
=(1,1,1,1,1,1,1,1,1,0)
\end{array}
$$

$g\left(x_{1}, x_{2}, \cdots, x_{10}\right)=0$, there is 1 such case.
Thus, the number of elements in the set $D_{10}(g)$ is
$$
\begin{array}{l}
256+64+16+4+1=341 . \\
\text { Therefore, } \sum_{\left(x_{1}, x_{2}, \cdots, x_{10}\right) \in D_{10}(g)}\left(x_{1}+x_{2}+\cdots+x_{10}\right) \\
=1 \times 256+128 \times 8+3 \times 64+32 \times 6+ \\
\quad 5 \times 16+8 \times 4+7 \times 4+2 \times 2+9 \\
=1817 .
\end{array}
$$