28. 已知集合A={x∣x2−3x+2⩽0}A=\{x\vert x^{2}-3x+2\leqslant 0\}A={x∣x2−3x+2⩽0},函数f(x)=x2−2ax+1f(x)=x^{2}-2ax+1f(x)=x2−2ax+1.(1)当a≠0a\ne 0a=0时,解关于xxx的不等式f(x)⩽3a2+1f(x)\leqslant 3a^{2}+1f(x)⩽3a2+1;(2)若命题"存在x0∈Ax_{0}\in Ax0∈A,使得f(x0)⩽0f(x_{0})\leqslant 0f(x0)⩽0"为假命题,求实数aaa的取值范围.