解(1)由f(2)=0,得4a+2b=0.

　　由方程f(x)=x,得ax2+(b-1)x=0.

　　因为方程f(x)=x有两个相等的实根,所以Δ=(b-1)2=0.解方程组{■(4a+2b=0"," @"(" b"-" 1")" ^2=0"," )┤得{■(a="-" 1/2 "," @b=1"," )┤

　　故f(x)=-1/2x2+x.

　　(2)由(1)知,f(x)=-1/2x2+x=-1/2(x-1)2+1/2≤1/2,即2n≤1/2,解得n≤1/4.故函数f(x)在[m,n]上是增函数.由{■(f"(" m")" ="-" 1/2 m^2+m=2m"," @f"(" n")" ="-" 1/2 n^2+n=2n"," )┤

　　解得m=-2或m=0,n=-2或n=0.

　　因为m