数列\(\{{{a}_{n}}\}\)的前\(n\)项和为\({{S}_{n}}\)，且\({{a}_{n}}\)是\({{S}_{n}}\)和\(1\)的等差中项，等差数列\(\{{{b}_{n}}\}\)满足\({{b}_{1}}={{a}_{1}}\)，\({{b}_{4}}={{S}_{3}}\)．

\((\)Ⅰ\()\)求数列\(\{{{a}_{n}}\}\)、\(\{{{b}_{n}}\}\)的通项公式；

\((\)Ⅱ\()\)设\({{c}_{n}}=\dfrac{1}{{{b}_{n}}{{b}_{n+1}}}\)，数列\(\{{{c}_{n}}\}\)的前\(n\)项和为\({{T}_{n}}\)，证明：\(\dfrac{1}{3}\leqslant {{T}_{n}} < \dfrac{1}{2}\)．