1. We claim that if $n \in A$, $n$ is even, then $6n \in A$, and $8n \in A$. (1) $n \in A \Rightarrow \frac{3n}{2}, 3n \notin A \Rightarrow \frac{9n}{2} \in A \Rightarrow 9n \notin A$. So $3n, 9n \notin A \Rightarrow 6n \in A$. (2) $6n=(2n)\cdot3 \in A \Rightarrow (2n)\cdot2=4n \notin A$; $6n \in A \Rightarrow (6n)\cdot2=12n=(4n)\cdot3 \notin A$. So $(4n)\cdot2=8n \in A$ 2. $2 \in A \Rightarrow 2 \cdot 8 \cdot 8=128 \in A \Rightarrow 128 \cdot 6 \cdot 6 \cdot 6=27648 \in A$, so that $\frac{27648}{2}= \boxed{13824 \notin A}$