已知数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)满足：\(S_{n}= \dfrac {a}{a-1}(a_{n}-1)(a\)为常数，且\(a\neq 0\)，\(a\neq 1)\)
\((1)\)若\(a=2\)，求数列\(\{a_{n}\}\)的通项公式
\((2)\)设\(b_{n}= \dfrac {2S_{n}}{a_{n}}+1\)，若数列\(\{b_{n}\}\)为等比数列，求\(a\)的值．
\((3)\)在满足条件\((2)\)的情形下，设\(c_{n}= \dfrac {1}{1+a_{n}}+ \dfrac {1}{1-a_{n+1}}\)，数列\(\{c_{n}\}\)前\(n\)项和为\(T_{n}\)，求证\(T_{n} > 2n- \dfrac {1}{3}\)．