4．（2023•东莞市校级三模）已知函数f(x)=1−cosxf(x)=1-\cos xf(x)=1−cosx．（1）证明：f(x)⩽x22f(x)\leqslant \frac{x^{2}}{2}f(x)⩽2x2；（2）证明：函数h(x)=aln(x+1)−f(x)(0<a<1)h(x)=aln(x+1)-f(x)(0<a<1)h(x)=aln(x+1)−f(x)(0<a<1)在(0,π2)(0,\frac{\pi }{2})(0,2π)上有唯一零点x0x_{0}x0，且x0>4a+1−1x_{0}>\sqrt{4a+1}-1x0>4a+1−1．