1. The domain of the function $y=\arccos \left(\frac{\sqrt{12+4 x-x^{2}}-2}{4}\right)$ is $\qquad$ , and the range is $\qquad$ .
$$
-1 .[-2,6],\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right] \text {. }
$$

From $12+4 x-x^{2} \geqslant 0$, we get $-2 \leqslant x \leqslant 6$. At this point,
$$
\begin{array}{l}
0 \leqslant \sqrt{12+4 x-x^{2}}=\sqrt{-(x-2)^{2}+16} \leqslant 4 \\
\Rightarrow-\frac{1}{2} \leqslant \frac{\sqrt{12+4 x-x^{2}}-2}{4} \leqslant \frac{1}{2} .
\end{array}
$$

Therefore, the domain is $[-2,6]$, and the range is
$$
\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right] \text {. }
$$