数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，已知\(S_{n+1}=S_{n}+a_{n}+2\)，\(a_{1}\)，\(a_{2}\)，\(a_{5}\)成等比数列．
\((1)\)求数列\(\{a_{n}\}\)的通项公式；
\((2)\)若数列\(\{b_{n}\}\)满足\( \dfrac {b_{n}}{a_{n}}=( \sqrt {2})^{1+a_{n}}\)，求数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}\)．