12. The product of all elements of a finite set $S$ is called the "product number" of the number set $S$. Now, given the number set $M=$ $\left\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \cdots, \frac{1}{100}\right\}$, determine the sum of the "product numbers" of all even-numbered (2, 4, $\cdots$, 98) element subsets of $M$.
12. Set $M$ contains 99 elements. Let the sum of the "product numbers" of all even-numbered element subsets of $M$ be $G$, and the sum of the "product numbers" of all odd-numbered element subsets be $H$. Then we have
$$
\begin{aligned}
G+H & =\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right) \cdots\left(1+\frac{1}{99}\right)\left(1+\frac{1}{100}\right)-1 \\
& =\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{100}{99} \cdot \frac{101}{100}-1=\frac{101}{2}-1=\frac{99}{2} \\
G-H & =\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \cdots\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right)-1 \\
& =\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{98}{99} \cdot \frac{99}{100}-1=\frac{1}{100}-1=-\frac{99}{100}
\end{aligned}
$$

Thus, $G=\frac{1}{2}\left(\frac{99}{2}-\frac{99}{100}\right)=24.255$.