$证明：(1)∵b²-4ac=[-(2m-1)]²-4×1× (-3m²+m)=16m²-8m+1=(4m-1)²≥0$
$∴无论m 为何值，该方程总有实数根$
$(2)由一元二次方程的根与系数的关系，得x_{1}+x_{2}=2m-1，x_{1}x_{2}=-3m²+m$
$∵\frac{x_{2}}{x_{1}}+\frac{x_{1}}{x_{2}}=- \frac{5}{2}$
$∴\frac{x_{1}^2+x_{2}^2}{x_{1}x_{2}}=- \frac{5}{2}，即 \frac{(x_{1}+x_{2})²-2x_{1}x_{2}}{x_{1}x_{2}}=- \frac{5}{2}，∴\frac{(x_{1}+x_{2})^2}{x_{1}x_{2}}= -\frac{1}{2}$
$∴\frac{(2m-1)^2}{-3m²+m}=-\frac{1}{2}，解得 m=1 或\frac{2}{5}$
$故m 的值为1或\frac{2}{5}$
