(15) 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 different ways to paint, then the range of the real number $m$ is $\qquad$.
(15) Clearly, the two fixed lines $y= \pm \sqrt{2} x$ divide the elliptical surface $\frac{x^{2}}{4}+y^{2} \leqslant 1$ into 4 parts (as shown in the figure). When $|m| \geqslant 2$, using 6 colors to paint these 4 parts, the number of different painting methods is $\mathrm{A}_{6}^{4}<720$, which does not meet the requirement, so $|m|<2$. By
$$
\left\{\begin{array}{l}
y= \pm \sqrt{2} x \\
\frac{x^{2}}{4}+y^{2}=1
\end{array}\right.
$$

we get $x= \pm \frac{2}{3}$.
When $m=0$, the line $x=0$ passes through the origin, at this time the elliptical surface is divided into 6 parts. Using 6 colors to paint these 6 parts, the number of different painting methods is $\mathrm{A}_{6}^{6}=720$ (ways), which meets the requirement.
When $0<|m|<\frac{2}{3}$, the line $x=m$ divides the elliptical surface into 7 parts. Using 6 colors to paint 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 paint these 5 parts, the number of different painting methods is $\mathrm{A}_{6}^{5}=720$ (ways). This meets the requirement.
In summary, the range of values for $m$ is
$$
\left\{m \left\lvert\,-2<m \leqslant-\frac{2}{3}\right. \text { or } m=0 \text { or } \frac{2}{3} \leqslant m<2\right\} \text {. }
$$