若数列\(\{a _{n} \}\)满足前\(n\)项之和\(S _{n} =2a _{n} -4(n∈N ^{*} )\)，\(b _{n+1} =a _{n} +2b _{n}\)，且\(b _{1} =2\)，
\((1)\)求数列\(\{a _{n} \}\)的通项公式
\((2)\)证明：\(\{ \dfrac {b_{n}}{2^{n}} \}\)是等差数列
\((3)\)求\(b _{n}\)的前\(n\)项和\(T _{n}\)．