已知等差数列\(\{a_{n}\}\)满足：\(a_{2}=5\)，\(a_{4}+a_{6}=22\)，\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)
\((1)\)求\(a_{n}\)及\(S_{n}\)；
\((2)\)令\(b_{n}= \dfrac {1}{a_{n}^{2}-1}(n∈N^{*})\)，求数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}\)．