2. Let complex numbers $z, w$ satisfy $|z|=3,(z+\bar{w})(\bar{z}-w)=7+4 \mathrm{i}$, where $\mathrm{i}$ is the imaginary unit, $\bar{z}, \bar{w}$ represent the conjugates of $z, w$ respectively, then the modulus of $(z+2 \bar{w})(\bar{z}-2 w)$ is $\qquad$ .
Answer: $\sqrt{65}$.
Solution: By the properties of operations, $7+4 \mathrm{i}=(z+\bar{w})(\bar{z}-w)=|z|^{2}-|w|^{2}-(z w-\overline{z w})$, since $|z|^{2}$ and $|w|^{2}$ are real numbers, $\operatorname{Re}(z w-\overline{z w})=0$, hence $|z|^{2}-|w|^{2}=7, z w-\overline{z w}=-4 \mathrm{i}$, and given $|z|=3$, so $|w|^{2}=2$. Therefore,
$$
(z+2 \bar{w})(\bar{z}-2 w)=|z|^{2}-4|w|^{2}-2(z w-\overline{z w})=9-8+8 \mathrm{i}=1+8 \mathrm{i} .
$$

Thus, the modulus of $(z+2 \bar{w})(\bar{z}-2 w)$ is $\sqrt{1^{2}+8^{2}}=\sqrt{65}$.