1. The range of the function $y=\sqrt{3 x+6}+\sqrt{8-x}$ is
1. $y \in[\sqrt{10}, 2\sqrt{10}]$.
The original function can be transformed into
$$
\begin{array}{l}
y=\sqrt{3} \sqrt{x+2}+ \\
\sqrt{8-x}.
\end{array}
$$

Let point $P(u, v)=P(\sqrt{x+2}, \sqrt{8-x})$, then
$$
u^{2}+v^{2}=10(u, v \geq 0),
$$

which means $P$ lies on the circle $u^{2}+v^{2}=10(u, v \geqslant 0)$. Also, $y=\sqrt{3} u+v$, which means $(u, v)$ lies on the line $v=-\sqrt{3} u+y$.

When the line $v=-\sqrt{3} u+y(u, v \geqslant 0)$ intersects with the arc $u^{2}+v^{2}=10(u, v \geqslant 0)$, the range of the intercept $y$ is the range of the function. It is easy to find that
$$
\sqrt{10} \leqslant y \leqslant 2 \sqrt{10}.
$$

Another solution: From $-2 \leqslant x \leqslant 8$, we can set $x=-2+10 \sin ^{2} x, x \in \left[0, \frac{\pi}{2}\right]$. Thus, $y=\sqrt{10}(\sqrt{3} \sin x+\cos x)$
$$
=2 \sqrt{10} \sin \left(x+\frac{\pi}{6}\right).
$$

Since $\frac{\pi}{6} \leqslant x+\frac{\pi}{6} \leqslant \frac{2 \pi}{3}$, we have $\frac{1}{2} \leqslant \sin \left(x+\frac{\pi}{6}\right) \leqslant 1$. Therefore, $\sqrt{10} \leqslant y \leqslant 2 \sqrt{10}$.