2. Let the complex numbers $z, w$ satisfy
$$
|z|=3,(z+\bar{w})(\bar{z}-w)=7+4 \mathrm{i} \text{, }
$$

where $\mathrm{i}$ is the imaginary unit, and $\bar{z}, \bar{w}$ are the conjugates of $z, w$ respectively. Then the modulus of $(z+2 \bar{w})(\bar{z}-2 w)$ is $\qquad$
2. $\sqrt{65}$.

From the properties of operations, we know
$$
\begin{array}{l}
7+4 \mathrm{i}=(z+\bar{w})(\bar{z}-w) \\
=|z|^{2}-|w|^{2}-(z w-\overline{z w}) .
\end{array}
$$
$$
\begin{array}{l}
\text { Since }|z|^{2}, |w|^{2} \in \mathbf{R}, \operatorname{Re}(z w-\overline{z w})=0 \\
\Rightarrow|z|^{2}-|w|^{2}=7, z w-\overline{z w}=-4 \mathrm{i} . \\
\text { Also }|z|=3 \Rightarrow|w|^{2}=2 .
\end{array}
$$

Then $(z+2 \bar{w})(\bar{z}-2 w)$
$$
\begin{array}{l}
=|z|^{2}-4|w|^{2}-2(z w-\overline{z w}) \\
=9-8+8 \mathrm{i}=1+8 \mathrm{i} .
\end{array}
$$

Therefore, the modulus of $(z+2 \bar{w})(\bar{z}-2 w)$ is $\sqrt{65}$.