等差数列\(\{a_{n}\}\)的各项均为正数，\(a_{1}=1\)，前\(n\)项和为\(S_{n}\)；数列\(\{b_{n}\}\)为等比数列，\(b_{1}=1\)，且\(b_{2}S_{2}=6\)，\(b_{2}+S_{3}=8\)．

\((1)\)求数列\(\{a_{n}\}\)与\(\{b_{n}\}\)的通项公式；

\((2)\)求\(\dfrac{1}{{{S}_{1}}}+\dfrac{1}{{{S}_{2}}}+\cdots +\dfrac{1}{{{S}_{n}}}\)．