$x_1<1<x_2$
$⇒ \left\{ \begin{array}{l}x_1-1<0\\x_2-1>0\end{array} \right.$
$⇒ (x_1-1)(x_2-1)<0$
$⇔ x_1x_2-(x_1+x_2)+1<0$
Theo định lý Vi-ét, ta có:
$\left\{ \begin{array}{l}x_1+x_2=2\\x_1x_2=3m-2\end{array} \right.$
Vậy $x_1x_2-(x_1+x_2)+1<0$
$⇔ 3m-2-2+1<0$
$⇔ 3m-3<0$
$⇔ 3m<3$
$⇔ m<1$
Vậy $m<1$.