a) Taking the inverse, $\frac{1}{(a_{n+1})} = \frac{((n+\frac{1}{2})a_n + 2^n)}{(2^{n+1}*a_n)} = \frac{(n + \frac{1}{2})}{(2^{n+1})} + \frac{1}{2(a_n)}$ Therefore, $\frac{1}{(a_{n+1})} - \frac{1}{2(a_n)} = \frac{(n+\frac{1}{2})}{(2^{n+1})}$ $\frac{1}{2(a_{n})} - \frac{1}{4(a_{n-1})} = \frac{(n-\frac{1}{2})}{(2^{n-1})}$ and so on, until $\frac{1}{(2^{n-1}*a_{2})} - \frac{1}{2^n*(a_1)} = \frac{(\frac{3}{2})}{(2^{3-n})}$ Summing it up, you get the general term for $a_{n+1}$, and then you can easily find the solution. Somebody please check if I made a mistake somewhere, I'm sleepy.

Good so far. Simplify a bit: $\frac{1}{a_n}-\frac{1}{2a_{n-1}}=\frac{1}{2^{n+1}}\left(n-\frac{1}{2}\right)$ Now sum: $\frac{1}{a_n}=\frac{1}{2^{n-1}a_1}+\sum_{k=2}^n\frac{1}{2^{n-k}}\left(\frac{1}{a_k}-\frac{1}{2a_{k-1}}\right)=\frac{1}{2^n}+\frac{1}{2^{n+1}}\sum_{k=2}^n(2k-1)=\frac{1}{2^n}+\frac{n^2-1}{2^{n+1}}=\frac{n^2+1}{2^{n+1}}$. Therefore $a_n=\frac{2^{n+1}}{n^2+1}$. Now try part (b).