8.(2023春•东城区校级月考)设函数f(x)=x(x2−3x+a)f(x)=x(x^{2}-3x+a)f(x)=x(x2−3x+a),a∈Ra\in Ra∈R.(1)当a=−9a=-9a=−9时,求函数f(x)f(x)f(x)的单调增区间;(2)若函数f(x)f(x)f(x)在区间(1,2)(1,2)(1,2)上为减函数,求aaa的取值范围;(3)若函数在区间(0,2)(0,2)(0,2)内存在两个极值点x1x_{1}x1,x2x_{2}x2,且∣f(x1)−f(x2)∣>∣f(x1)+f(x2)∣\vert f(x_{1})-f(x_{2})\vert >\vert f(x_{1})+f(x_{2})\vert∣f(x1)−f(x2)∣>∣f(x1)+f(x2)∣,求aaa的取值范围.