(1)方程有两个实根,∴2^2-4m≥0 解得m≤1(2)x2/x1+x1/x2=(x2²+x1²)/(x1x2)=[(x1+x2)²-2x1x2]/(x1x2)=[4-2m]/m=1解得m=4/3∵4/3>1,∴不存在