25 Two equilateral triangles overlap to form a six-pointed star (as shown in the figure). The first 12 positive integers 1, $2, \cdots, 12$ are to be placed at the 12 nodes of the figure, such that the sum of the four numbers on each straight line is equal.
(1) Find the minimum sum of the numbers placed at the six vertices $a_{1}$, $a_{2}, \cdots, a_{6}$ of the six-pointed star;
(2) Prove that the number of different filling schemes that satisfy the condition is even. (For filling schemes $\pi$ and $T$, if scheme $\pi$ can be rotated or flipped to coincide with scheme $T$, they are considered the same filling scheme)
(1) For any filling method that satisfies the conditions, the number filled at point $a_{i}$ is still denoted as $a_{i}, i=1,2, \cdots, 12$. If the sum of the four numbers on each line is $s$, then from $6 s=2(1+2+\cdots+12)$, we get $s=26$;
In $\triangle a_{1} a_{3} a_{5}$, on the three sides, we have
$$
\begin{array}{l}
\left(a_{1}+a_{8}+a_{9}+a_{3}\right)+\left(a_{3}+a_{10}+a_{11}+a_{5}\right) \\
\quad+\left(a_{5}+a_{12}+a_{7}+a_{1}\right)=3 s ;
\end{array}
$$

In $\triangle a z a_{4} a_{6}$, on the three sides, we have
$$
\begin{array}{l}
\left(a_{2}+a_{9}+a_{10}+a_{4}\right)+\left(a_{4}+a_{11}+a_{12}+a_{6}\right) \\
\quad+\left(a_{6}+a_{7}+a_{8}+a_{2}\right)=3 s .
\end{array}
$$

Subtracting the two equations, we get $a_{1}+a_{3}+a_{5}=a_{2}+a_{4}+a_{6}$. 
From this, we know that the sum of the numbers at the vertices of the two triangles is equal. If we denote $m=\left(a_{1}+a_{3}+\right.$ $\left.a_{5}\right)+\left(a_{2}+a_{4}+a_{6}\right)$, then $m$ is an even number.

Because in $1,2, \cdots, 12$, the sum of the smallest six numbers $1+2+\cdots+6=21$ is odd, then $m \geqslant 22$. If $m=22$, since in $1,2, \cdots, 12$, the sum of six numbers that equals 22 is only $1,2,3,4,5,7$, dividing these into two groups with equal sums, there is only one unique division:
$$
\{1,3,7\} \text { and }\{2,4,5\} \text {. }
$$

Now, ignoring the specific numbers on the outer circle, and changing $\{1,3,7\}$ and $\{2,4,5\}$ to $\{$ odd, odd, odd $\}$ and $\{$ even, even, odd $\}$, respectively, then under rotation, the filling method of the six vertices is unique.

Now consider the filling method of the six inner numbers. Using the notation $a \sim b$ to indicate that integers $a, b$ have the same parity, since the sum of the four numbers on each line $s$ is even, then $a_{9} \sim a_{8} \sim a_{7} \sim a_{12}, a_{10} \sim$ $a_{11}$, and in the six inner numbers $6, 8, 9, 10, 11, 12$, there are exactly four even numbers and two odd numbers, so $\left\{a_{10}, a_{11}\right\}=\{9,11\}$. Since $a_{10}, a_{11}$ are on one side of the triangle with odd vertices, the sum of the numbers at the two vertices of that side should be $s-(9+11) = 6$, but in $\{1,3,7\}$, there are no two numbers that sum to 6, which is a contradiction. Therefore, $m \neq 22$, and thus the even number $m \geqslant 24$.

When $m=24$, we can actually provide a filling method that meets the requirements (as shown in the figure), so the minimum value sought is 24.
(2) For any filling method $\pi$ that satisfies the conditions, let the numbers at each point be $a_{1}, a_{2}, \cdots, a_{12}$, and let
$$
b_{i}=13-a_{i}, i=1,2, \cdots, 12,
$$

then when $a_{i} \in\{1,2, \cdots, 12\}$, $b_{i} \in\{1,2, \cdots, 12\}$. Now, in the filling method $\pi$, replace $a_{1}, a_{2}, \cdots, a_{12}$ with $b_{1}, b_{2}, \cdots, b_{12}$, to get the filling method $T$, then $T$ is also a filling method that satisfies the conditions; obviously
$$
\sum_{i=1}^{6} a_{i}+\sum_{i=1}^{6} b_{i}=\sum_{i=1}^{6}\left(a_{i}+b_{i}\right)=78,
$$

and $\sum_{i=1}^{6} a_{i}, \sum_{i=1}^{6} b_{i}$ are both even numbers, so $\sum_{i=1}^{6} a_{i} \neq 39, \sum_{i=1}^{6} b_{i} \neq 39$.
Let the set of all filling methods that satisfy the conditions be $M$, and divide $M$ into two disjoint subsets: $M=M_{1} \cup M_{2}$, where, in $M_{1}$, each filling method $\pi_{1}=\left(a_{1}, a_{2}, \cdots\right.$, $a_{12}$ ), the sum of the numbers at the six vertices $\sum_{i=1}^{6} a_{i}39$.

Make a correspondence between the filling method $\pi_{1}=\left(a_{1}, a_{2}, \cdots, a_{12}\right)$ in $M_{1}$ and the filling method $\pi_{2}=$ $\left(b_{1}, b_{2}, \cdots, b_{12}\right)$ in $M_{2}$, such that $b_{i}=13-a_{i}, i=1,2, \cdots, 12$, this correspondence is one-to-one, therefore, $\left|M_{1}\right|=\left|M_{2}\right|, M_{1} \cap M_{2}=\varnothing$, and $|M|=$ $\left|M_{1} \cup M_{2}\right|=2\left|M_{1}\right|$, i.e., the number of elements in $M$ is even. That is, the number of different filling methods that satisfy the conditions is even.