已知首项是\(5\)的数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，且\(S_{n+1}+1=S_{n}+2a_{n}+2^{n+1}\)，数列\(\{b_{n}\}\)满足\(b_{n}=\dfrac{a_{n}-1}{2^{n}}.\)
\((Ⅰ)\)证明：\(\{b_{n}\}\)是等差数列，并求\(\{b_{n}\}\)的通项公式；
\((Ⅱ)\)求数列\(\{a_{2n}\}\)的前\(n\)项和\(T_{n}.\)