The last three digits of $ 2^x$ must form a number divisible by 8 and its last digit is nonzero, implying that these three digits can take only values from the following set $ M=\{ 008, 152, 224, 512\}$. Consider the congruence $ 2^x\equiv m\pmod{1000}$ that is equivalent to $ 2^{x-3}\equiv m/8\pmod{125}$ where $ m\in M$ and thus $ m/8\in \{ 1, 19, 28, 64 \}$. For each of them, the congruence has one solution modulo 125. Therefore, the total number of solutions modulo 125 is 4, implying that \[ 4\lfloor \frac{n-6}{125} \rfloor \leq n(A) \leq 4\lceil \frac{n-6}{125} \rceil.\] Since $ n(S)=n-6$, we have \[ \frac{4}{125} - \frac{4}{n-6} \rfloor \leq \frac{n(A)}{n(S)} \leq \frac{4}{125} + \frac{4}{n-6}.\] Together with $ n\geq 2009$ the latter inequality gives the required bounds for $ \frac{n(A)}{n(S)}.$