4.(2023春•渝中区校级期末)(1)不等式lnx⩽mx−1lnx\leqslant mx-1lnx⩽mx−1对任意的x>0x>0x>0恒成立,求mmm的取值范围.(2)当a∈(0,1)a\in (0,1)a∈(0,1),求证:ex−x2+(a−13)x>lnx{e^x}-{x^2}+({a-\frac{1}{3}})x>lnxex−x2+(a−31)x>lnx(参考数据:e2≈7.4e^{2}\approx 7.4e2≈7.4,e3≈20.1)e^{3}\approx 20.1)e3≈20.1).