7. 若 $\tan \alpha+\frac{1}{\tan \alpha}=3$, 则 $\sin \alpha \cos \alpha=$ \$ \qquad \$ $\tan ^{2} \alpha+\frac{1}{\tan ^{2} \alpha}=$ \$ \qquad \$ .
because \tan \alpha+\frac{1}{\tan \alpha}=\frac{1}{\cos \alpha \sin \alpha}=3$,

$\therefore \sin \alpha \cos \alpha=\frac{1}{3}$.

又 $\tan ^{2} \alpha+\frac{1}{\tan ^{2} \alpha}=\left(\tan \alpha+\frac{1}{\tan \alpha}\right)^{2}-2=9-2=7$,

$\therefore \tan ^{2} \alpha+\frac{1}{\tan ^{2} \alpha}=7$.

【答案】 $\frac{1}{3} \quad 7$