8. Let $A, B, C$ be the three interior angles of $\triangle ABC$, then the range of values for $3 \cos A + 2 \cos 2B + \cos 3C$ is $\qquad$
Let $A \rightarrow 0, B \rightarrow \pi, C \rightarrow 0 \Rightarrow 3 \cos A + 2 \cos 2B + \cos 3C \rightarrow 6$;
Let $\cos A \rightarrow -\frac{3}{8}, \cos B \rightarrow \frac{3}{8}, C \rightarrow 0$ when, $\cos 2B = 2 \times \frac{9}{64} - 1 = -\frac{23}{32}$, then $3 \cos A + 2 \cos 2B + \cos 3C \rightarrow -\frac{9}{8} - \frac{23}{16} + 1 = -\frac{25}{16}$.
Clearly, $3 \cos A + 2 \cos 2B + \cos 3C - \frac{25}{16}$. If $A \leqslant \frac{\pi}{3}$, then $3 \cos A + 2 \cos 2B + \cos 3C \geqslant \frac{3}{2} - 2 - 1 = -\frac{3}{2} > -\frac{25}{16}$;
If $B \leqslant C$, then $A + 2B \leqslant A + B + C = \pi \Rightarrow A \leqslant \pi - 2B \Rightarrow \cos A + \cos 2B \geqslant 0$, $A + C - \cos C$,
Thus, $3 \cos A + 2 \cos 2B + \cos 3C \geqslant \cos A + \cos 3C > \cos 3C - \cos C$
$$
\begin{array}{l}
= -2 \sin 2C \sin C = -4 \sin^2 C \cos C = -2 \sqrt{2} \cdot \sqrt{\sin^2 C \cdot \sin^2 C \cdot (2 - 2 \sin^2 C)} \\
\geqslant -2 \sqrt{2} \cdot \sqrt{\left(\frac{2}{3}\right)^3} = -\frac{8}{3 \sqrt{3}} > -\frac{25}{16} ;
\end{array}
$$

If $A > \frac{\pi}{3}$, and $B > C$, since $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cdot \cos \frac{A-B}{2}$,
$$
\cos (A + B - C) + \cos C = 2 \cos \frac{A+B}{2} \cdot \cos \frac{A+B-2C}{2},
$$

then $\cos \frac{A-B}{2} > \cos \frac{A+B-2C}{2} \Rightarrow \cos A + \cos B > \cos (A + B - C) + \cos C$.
Thus, $3(\cos A + \cos B) + \cos 3C > 3[\cos (A + B - C) + \cos C] + \cos 3C$
$$
\begin{array}{l}
= 3[\cos (\pi - 2C) + \cos C] + 4 \cos^3 C - 3 \cos C = -3 \cos 2C + 4 \cos^3 C \\
= -3(2 \cos^2 C - 1) + 4 \cos^3 C = 1 + 2(2 \cos^3 C - 3 \cos^2 C + 1) \\
= 1 + 2(2 \cos C + 1)(\cos C - 1)^2 > 1 \Rightarrow 3 \cos A + \cos 3C > 1 - 3 \cos B .
\end{array}
$$

Therefore, $3 \cos A + 2 \cos 2B + \cos 3C > 2 \cos 2B - 3 \cos B + 1 = 4 \cos^2 B - 3 \cos B - 1$
$$
= 4\left(\cos B - \frac{3}{8}\right)^2 - \frac{25}{16} \geqslant -\frac{25}{16} \text{. }
$$

In summary, the range of $3 \cos A + 2 \cos 2B + \cos 3C$ is $\left(-\frac{25}{16}, 6\right)$.