6. Let $m>1$, under the constraint conditions
$$
\left\{\begin{array}{l}
y \geqslant x, \\
y \leqslant m x, \\
x+y \leqslant 1
\end{array}\right.
$$

the maximum value of the objective function $z=x+m y$ is less than 2. Then the range of values for $m$ is $\qquad$
6. $(1,1+\sqrt{2})$.

The objective function is transformed into $y=-\frac{1}{m} x+\frac{z}{m}$.
Given $m>1$, we know $-1<-\frac{1}{m}<0$.
The feasible region represented by the system of inequalities is the shaded area in Figure 2.
Figure 2
According to the geometric meaning of the objective function, the maximum value of the objective function is achieved when the intercept of the line $y=-\frac{1}{m} x+\frac{z}{m}$ on the $y$-axis is the largest.
Clearly, the maximum value is achieved at point $A$.
By solving $\left\{\begin{array}{l}y=m x, \\ x+y=1,\end{array}\right.$, we get the intersection point $A\left(\frac{1}{1+m}, \frac{m}{1+m}\right)$.
Thus, the maximum value of the objective function is
$$
\begin{array}{l}
\frac{1}{1+m}+\frac{m^{2}}{1+m}<2 \Rightarrow m^{2}-2 m-1<0 \\
\Rightarrow 1-\sqrt{2}<m<1+\sqrt{2} .
\end{array}
$$

Therefore, the range of $m$ is $(1,1+\sqrt{2})$.