8. The three lines $y=\sqrt{2} x, y=-\sqrt{2} x$ and $x=m$ divide the elliptical region $\frac{x^{2}}{4}+y^{2} \leqslant 1$ into several parts. Now, using 6 different colors to paint these parts, each part is painted with one color, and any two parts are of different colors, there are 720 ways to paint. Then the range of the real number $m$ is $\qquad$
8. $\left(-2,-\frac{2}{3}\right] \cup\{0\} \cup\left[\frac{2}{3}, 2\right)$.

Obviously, the two fixed lines $y= \pm \sqrt{2} x$ divide the elliptical region $\frac{x^{2}}{4}+y^{2} \leqslant 1$ into 4 parts. When $|m| \geqslant 2$, using 6 colors to color these 4 parts, the number of different coloring methods is $\mathrm{A}_{6}^{4}<720$, which does not meet the requirement, so $|m|<2$. By solving
$$
\left\{\begin{array}{l}
y= \pm \sqrt{2} x \\
\frac{x^{2}}{4}+y^{2}=1
\end{array}\right.
$$

we get $x= \pm \frac{2}{3}$.
When $m=0$, the line $x=0$ passes through the origin, at this time the elliptical region is divided into 6 parts, using 6 colors to color these 6 parts, the number of different coloring methods is $\mathrm{A}_{6}^{6}=720$, which meets the requirement.

When $0<|m|<\frac{2}{3}$, the line $x=m$ divides the elliptical region into 7 parts, using 6 colors to color these 7 parts, there must be two parts of the same color, which does not meet the requirement.

When $\frac{2}{3} \leqslant|m|<2$, the line $x=m$ divides one of the parts into two parts, resulting in a total of 5 parts. Using 6 colors to color these 5 parts, the number of different coloring methods is $\mathrm{A}_{6}^{5}=720$, which meets the requirement.
In summary, the range of values for $m$ is $\left(-2,-\frac{2}{3}\right] \cup\{0\} \cup\left[\frac{2}{3}, 2\right)$.