7.
已知\({数列}\{ a_{n}\}\)的前\(n\)项\({和}{为}A_{n}\),对\({任意}n{∈}N^{{*}}{满足}\dfrac{A_{n{+}1}}{n{+}1}{-}\dfrac{A_{n}}{n}{=}\dfrac{1}{2}{,}{且}a_{1}{=}1\),\({数列}\{ b_{n}\}{满}{足}b_{n{+}2}{-}2b_{n{+}1}{+}b_{n}{=}0(n{∈}N{*}){,}b_{3}{=}5\),其前\(9\)项和为\(63{.}(1)\)求\({数列}\{ a_{n}\}{和}\{ b_{n}\}\)的通项公式\({;}(2){令}c_{n}{=}\dfrac{b_{n}}{a_{n}}{+}\dfrac{a_{n}}{b_{n}}\),\({数列}\{ c_{n}\}\)的前\(n\)项\({和}{为}T_{n}\),若对任意正整数\(n\),\({都}{有}T_{n}{\geqslant }2n{+}a\),求实数\(a\)的取值范围\({;}(3)\)将\({数列}\{ a_{n}\}{,}\{ b_{n}\}\)的项按照“当\(n\)为奇数\({时}{,}a_{n}\)放在前面;当\(n\)为偶数\({时}{,}b_{n}\)放在前面”的要求进行“交叉排列”,得到一个新的数列:\({\ \ \ \ \ \ \ \ \ \ \ \ \ \ a}_{1}{,}b_{1}{,}b_{2}{,}a_{2}{,}a_{3}{,}b_{3}{,}b_{4}{,}a_{4}{,}a_{5}{,}b_{5}{,}b_{6}{,…}\),
求这个新数列的前\(n{项}{和}S_{n}\).