设正项等比数列\(\{a_{n}\}\)的公比\(q\in N^{*}\)，且\((a_{1}-1)a_{3}=8\)，\(a_{1}+a_{2}^{2}=18\)，设数列\(\{b_{n}\}\)满足\(b_{1}=1\)，\((b_{n+1}-1)(b_{n}+3)=-4.\)
\((1)\)求\(\{a_{n}\}\)的通项公式；
\((2)\)当\(n\geqslant 2\)时，求\(a_{2}b_{2}+\dfrac{a_{3}b_{3}}{2}+\dfrac{a_{4}b_{4}}{3}+\)…\(+\dfrac{a_{n+1}b_{n+1}}{n}.\)