Consider an arbitrary $x \in\{1,2\}$, and let us consider the number of ways for $x$ to be in some of the sets so that the constraints are satisfied. We divide into a few cases: - Case: $x \notin A_{1}$. Then $x$ cannot be in any of the sets. So there is one possibility. - Case: $x \in A_{1}$ but $x \notin A_{2}$. Then the only other sets that $x$ could be in are $A_{3}, A_{5}, A_{7}$, and $x$ could be in some collection of them. There are 8 possibilities in this case. - Case: $x \in A_{2}$. Then $x \in A_{1}$ automatically. There are 4 independent choices to be make here: (1) whether $x \in A_{5} ;(2)$ whether $x \in A_{7} ;(3)$ whether $x \in A_{3}$, and if yes, whether $x \in A_{6}$; (4) whether $x \in A_{4}$, and if yes, whether $x \in A_{8}$. There are $2 \times 2 \times 3 \times 3=36$ choices here. Therefore, there are $1+8+36=45$ ways to place $x$ into some of the sets. Since the choices for $x=1$ and $x=2$ are made independently, we see that the total number of possibilities is $45^{2}=2025$.