【Question 1】
From $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}$, remove two addends so that the sum of the remaining four addends is exactly 1. Then, the two addends removed are $\qquad$ and $\qquad$.
$$
\begin{array}{l}
\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{10}+\left(\frac{1}{6}+\frac{1}{12}\right)=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{10}+\frac{1}{4}=1+\frac{1}{8}+\frac{1}{10} \\
\text { and } \frac{1}{8}+\frac{1}{10}=\frac{9}{40}
\end{array}
$$
$40=2^{3} \times 5$, the denominator containing the factor 5 is only $\frac{1}{10}$, so the other number is $\frac{1}{8}$;
the two remaining addends to be deleted are $\frac{1}{8}$ and $\frac{1}{10}$.