2. Given $-\frac{\pi}{2}<\alpha<\frac{\pi}{2}, 2 \tan \beta=\tan 2 \alpha, \tan (\beta-\alpha)=-2 \sqrt{2}$. Then $\cos \alpha=$
2. $\frac{\sqrt{3}}{3}$.

Let $\tan \alpha=u$. From
$$
\tan \beta=\frac{1}{2} \tan 2 \alpha=\frac{\tan \alpha}{1-\tan ^{2} \alpha}=\frac{u}{1-u^{2}},
$$

we get
$$
\tan (\beta-\alpha)=\frac{\tan \beta-\tan \alpha}{1+\tan \alpha \cdot \tan \beta}=\frac{\frac{u}{1-u^{2}}-u}{1+u \cdot \frac{u}{1-u^{2}}}=u^{3} \text {. }
$$

From $u^{3}=-2 \sqrt{2}$, we get $u=\tan \alpha=-\sqrt{2}$. Since $-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$, we have
$$
\cos \alpha=\frac{1}{\sqrt{1+\tan ^{2} \alpha}}=\frac{\sqrt{3}}{3} .
$$