2. Given that the circumcenter of $\triangle A B C$ is $O$, and
$$
\overrightarrow{A O} \cdot \overrightarrow{B C}=3 \overrightarrow{B O} \cdot \overrightarrow{A C}+4 \overrightarrow{C O} \cdot \overrightarrow{B A} \text {. }
$$

Then the range of $\cos B$ is $\qquad$
2. $\left[\frac{\sqrt{2}}{3}, 1\right)$.

As shown in Figure 3, take the midpoint $D$ of $B C$, and connect $O D$ and $A D$.
Figure 3

Since the circumcenter of $\triangle A B C$ is $O$, we have
$O D \perp B C$.
$$
\begin{array}{l}
\text { By } \overrightarrow{A O} \cdot \overrightarrow{B C}=(\overrightarrow{A D}+\overrightarrow{D O}) \cdot \overrightarrow{B C} \\
=\overrightarrow{A D} \cdot \overrightarrow{B C}+\overrightarrow{D O} \cdot \overrightarrow{B C}=\overrightarrow{A D} \cdot \overrightarrow{B C}, \\
\overrightarrow{A D} \cdot \overrightarrow{B C}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot(\overrightarrow{A C}-\overrightarrow{A B}) \\
=\frac{1}{2}\left(|\overrightarrow{A C}|^{2}-|\overrightarrow{A B}|^{2}\right)=\frac{1}{2}\left(b^{2}-c^{2}\right)
\end{array}
$$

Thus, $\overrightarrow{A O} \cdot \overrightarrow{B C}=\frac{1}{2}\left(b^{2}-c^{2}\right)$.
Similarly, $\overrightarrow{B O} \cdot \overrightarrow{A C}=\frac{1}{2}\left(a^{2}-c^{2}\right)$,
$$
\begin{array}{l}
\overrightarrow{C O} \cdot \overrightarrow{B A}=\frac{1}{2}\left(b^{2}-a^{2}\right) . \\
\text { Therefore, } \overrightarrow{A O} \cdot \overrightarrow{B C}=3 \overrightarrow{B O} \cdot \overrightarrow{A C}+4 \overrightarrow{C O} \cdot \overrightarrow{B A} \\
\Rightarrow \frac{1}{2}\left(b^{2}-c^{2}\right) \\
\quad=3 \times \frac{1}{2}\left(a^{2}-c^{2}\right)+4 \times \frac{1}{2}\left(b^{2}-a^{2}\right) \\
\Rightarrow a^{2}+2 c^{2}=3 b^{2} .
\end{array}
$$

By the cosine rule,
$\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$
$$
\begin{array}{l}
=\frac{a^{2}+c^{2}-\frac{1}{3}\left(a^{2}+2 c^{2}\right)}{2 a c} \\
=\frac{1}{3} \times \frac{2 a^{2}+c^{2}}{2 a c} .
\end{array}
$$

Notice that,
$$
\frac{2 a^{2}+c^{2}}{2 a c} \geqslant \frac{2 \sqrt{2} a c}{2 a c}=\sqrt{2},
$$

with equality holding if and only if $\sqrt{2} a=c$.
Combining with $0<\angle B<\pi$, we get
$$
\frac{\sqrt{2}}{3} \leqslant \cos B<1 \text {. }
$$