2.
已知函数\(f(x)\)满足\(2f\left(x+2\right)=f\left(x\right), \)当\(x∈\left(0,2\right) \)时,\(f\left(x\right)=\ln x+ax\left(a < - \dfrac{1}{2}\right),x∈\left(-4,-2\right) \)时,的最大值为\(-4\).
\((\)Ⅰ\()\)求\(x∈\left(0,2\right) \)时函数\(f(x)\)的解析式;
\((\)Ⅱ\()\)是否存在实数\(b\)使得不等式\(\dfrac{x-b}{f\left(x\right)+x} > \sqrt{x} \)对于\(x∈\left(0,1\right)∪\left(1,2\right) \)时恒成立,若存在,求出实数\(b\)的取值范围\(;\)若不存在,说明理由.