(1) 证明：$\because\Delta =[-(2m - 1)]^2 - 4×1×(-3m^2 + m)=4m^2 - 4m + 1 + 12m^2 - 4m = 16m^2 - 8m + 1=(4m - 1)^2\geq0，$
$\therefore$无论$m$为何值，方程总有实数根。
(2) 解：由题意知，$x_1 + x_2 = 2m - 1，$$x_1x_2 = - 3m^2 + m。$
$\because\frac{x_2}{x_1}+\frac{x_1}{x_2}=\frac{x_1^2 + x_2^2}{x_1x_2}=\frac{(x_1 + x_2)^2}{x_1x_2}-2=-\frac{5}{2}，$
$\therefore\frac{(2m - 1)^2}{-3m^2 + m}-2=-\frac{5}{2}。$
方程两边同时乘以$2(-3m^2 + m)$去分母得：
$2(2m - 1)^2-4(-3m^2 + m)=-5(-3m^2 + m)，$
$2(4m^2 - 4m + 1)+12m^2 - 4m = 15m^2 - 5m，$
$8m^2 - 8m + 2 + 12m^2 - 4m = 15m^2 - 5m，$
$20m^2 - 12m + 2 = 15m^2 - 5m，$
移项合并得$5m^2 - 7m + 2 = 0，$
因式分解得$(5m - 2)(m - 1)=0，$
解得$m = 1$或$m=\frac{2}{5}。$
