Solution:

1. Let $B$ be a subset of $A$ having the given property. Since $1 + 3 = 2^{2}$, we have that exactly one of the numbers $1$ or $3$ belongs to $B$.
If $1 \in B$ then $3 \notin B$. We prove by induction that for any integer $t$, $0 \leq t < 2^{n-2}$, the integers of the form $4t+1$ belong to $B$ and the integers of the form $4t+3$ do not belong to $B$. The statement is true for $t=0$ and suppose it is true for $t \leq s$. Since $4(s+1)+1$ is an odd number there exists $l$ such that $2^{l} < 4(s+1)+1 < 2^{l+1}$. Therefore $2(4s+5) > 2 \cdot 2^{l} = 2^{l+1}$ giving $0 < 2^{l+1} - (4s+5) < 4s+5$. Set $x = 4s+5$ and $y = 2^{l+1} - (4s+5)$. Then $x + y = 2^{l+1}$ and since $y$ is of the form $4m+3$ we conclude that $y \notin B$ and therefore $4(s+1)+1 \in B$. Analogously $4(s+1)+3 \notin B$.
If $1 \notin B$ then $3 \in B$ and we prove as above that the integers of the form $4t+1$ belong to $B$ and the integers of the form $4t+3$ do not.
Therefore the odd numbers in $B$ are either all integers of the form $4t+1$ or all integers of the form $4t+3$.
Let $x = 2^{p} x_{0}$ and $y = 2^{q} y_{0}$, where $x_{0}$ and $y_{0}$ are odd and $p$ and $q$ are positive integers. If $2^{p} x_{0} + 2^{q} y_{0} = 2^{k}$ and $p \neq q$, say $p < q$, then $x_{0} + 2^{q-p} y_{0} = 2^{k-p}$, which is impossible. Therefore $p = q$ and it follows that the sum of the elements from distinct sets $A_{i} = \{2^{i} a : a$ is an odd integer $\}, i = 1, 2, \ldots, n$ is not a power of $2$. For any $A_{i}$, after dividing by $2^{i}$ and applying the above arguments, we obtain that either all integers of the form $2^{i}(4t+1)$ are in $B$ or all integers of the form $2^{i}(4t+3)$ are in $B$.
Therefore there exist $2^{n+1}$ sets $B$ with the given property.