$b_n=\dfrac{n^2+1}{2}$

由已知可得$$\dfrac{a_{n+1}}{2^{n+1}}=\dfrac{a_n}{\left(n+\frac12\right)a_n+2^n},$$所以$$\dfrac{2^{n+1}}{a_{n+1}}-\dfrac{2^n}{a_n}=n+\dfrac12,$$即$$b_{n+1}-b_n=n+\dfrac12,$$从而\[\begin{split}&b_2-b_1=1+\dfrac12,\\&b_3-b_2=2+\dfrac12,\\&\cdots,\\&b_n-b_{n-1}=(n-1)+\dfrac12,\end{split}\]累加得$$b_n-b_1=1+2+\cdots+(n-1)+\dfrac{n-1}{2}=\dfrac{n^2-1}{2}.$$又因为 $b_1=\dfrac{2}{a_1}=1$，所以$$b_n=\dfrac{n^2-1}{2}+1=\dfrac{n^2+1}{2}.$$

$\dfrac12\left[1-\left(\dfrac12\right)^{n+1}\cdot\dfrac{n+2}{n+1}\right]$

由 $(1)$ 知$$a_{n+1}=\dfrac{2^{n+2}}{(n+1)^2+1},$$所以\[\begin{split}c_n&=\dfrac{(n+1)^2+1}{n(n+1)2^{n+2}}\\&=\dfrac12\left[\dfrac{n^2+n}{n(n+1)2^{n+1}}+\dfrac{n+2}{n(n+1)2^{n+1}}\right]\\&=\dfrac12\left[\dfrac{1}{2^{n+1}}+\dfrac{1}{n\cdot 2^n}-\dfrac{1}{(n+1)2^{n+1}}\right].\end{split}\]因此\[\begin{split}S_n&=\dfrac12\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\cdots+\dfrac{1}{2^{n+1}}\right)+\dfrac12\left\{\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot2^2}\right)+\cdots+\left[\dfrac{1}{n\cdot2^n}-\dfrac{1}{(n+1)\cdot2^{n+1}}\right]\right\}\\&=\dfrac12\cdot\dfrac{\frac{1}{2^2}\left(1-\frac{1}{2^n}\right)}{1-\frac12}+\dfrac12\cdot\left[\dfrac12-\dfrac{1}{(n+1)\cdot2^{n+1}}\right]\\&=\dfrac12\cdot\left[1-\left(\dfrac12\right)^{n+1}\cdot\dfrac{n+2}{n+1}\right].\end{split}\]