Question 1: Let set $A=\{1,2, m\}$, where $m$ is a real number. Let $B=\left\{a^{2} \mid a \in A\right\}, C=A \cup B$. If the sum of all elements in $C$ is 6, then the product of all elements in $\mathrm{C}$ is $\qquad$
Question 1, Answer: According to the conditions, we know: $C=\{1,2, m\} \cup\left\{4, \mathrm{~m}^{2}\right\}$. Since $1+2+4+\mathrm{m}+\mathrm{m}^{2}>6$, $1+2+4+m^{2}>6$, it can only be that $C=\{1,2,4, m\}$, thus we know $m=-1$. Therefore, the product of all elements in $C$ is -8.