22.(2023春•昆明期末)已知函数f(x)=ax3+bx2+1(a,b∈R)f(x)=ax^{3}+bx^{2}+1(a,b\in R)f(x)=ax3+bx2+1(a,b∈R)在x=1x=1x=1处取得极值0.(1)求aaa,bbb;(2)若过点(1,m)(1,m)(1,m)存在三条直线与曲线y=f(x)y=f(x)y=f(x)相切,求实数mmm的取值范围.