13. Given the equation about $x$
$$
8^{x}-m 4^{x}+\left(m^{2}-3\right) 2^{x}=0(m \in \mathbf{R}) \text {. }
$$
(1) If the equation has no real roots, find the range of $m$;
(2) If the equation has a unique real root, find the range of $m$.
13. Let $t=2^{x}$, then the original equation is transformed into an equation about $t$:
$$
t^{2}-m t+m^{2}-3=0 \text {. }
$$

Then: $\Delta=12-3 m^{2}$.
(1) The original equation has no real roots if and only if equation (1) has no real roots or only non-positive roots.
When equation (1) has no real roots, $m2$.
When equation (1) has only non-positive roots, $-2 \leqslant m \leqslant 2$.
By Vieta's formulas, we have $\left\{\begin{array}{l}m^{2}-3 \geqslant 0, \\ m \leqslant 0,\end{array}\right.$ which gives
$$
-2 \leqslant m \leqslant-\sqrt{3} \text {. }
$$

In summary, when the original equation has no real roots,
$$
m \leqslant-\sqrt{3} \text { or } m>2 \text {. }
$$
(2) The original equation has a unique real root if and only if equation (1) has two identical positive roots or one positive and one non-positive root. The original equation has two distinct real roots if and only if equation (1) has two distinct positive roots.
When equation (1) has two distinct positive roots, by Vieta's formulas, we have
$$
\left\{\begin{array}{l}
-20, \\
m>0,
\end{array} \text { which gives } \sqrt{3}<m<2 .\right.
$$

Since the original equation has at most two distinct real roots, when the original equation has only a unique real root, $m=2$ or $-\sqrt{3}<m \leqslant \sqrt{3}$.