已知数列\(\{ a_{n}\}{,}\{ b_{n}\}\)满足\(a_{1}{=}\dfrac{1}{4}{,}a_{n}{+}b_{n}{=}1{,}b_{n{+}1}{=}\dfrac{b_{n}}{1{-}a_{n}^{2}}\)．
\((1)\)求\(b_{1}{,}b_{2}{,}b_{3}{,}b_{4}\)；
\((2)\)求证：数列\(\{\dfrac{1}{b_{n}{-}1}\}\)是等差数列，并求出数列\(\{ b_{n}\}\)通项公式；