7. Let $H$ be the orthocenter of $\triangle A B C$, and
$$
3 \overrightarrow{H A}+4 \overrightarrow{H B}+5 \overrightarrow{H C}=\mathbf{0}
$$

Then $\cos \angle A H B=$ $\qquad$
7. $-\frac{\sqrt{6}}{6}$.

From the given, we have $\frac{\tan A}{3}=\frac{\tan B}{4}=\frac{\tan C}{5}=\lambda$.
Also, from $\tan A+\tan B+\tan C$
$$
=\tan A \cdot \tan B \cdot \tan C,
$$

we get $\lambda=\frac{1}{\sqrt{5}}, \tan C=\sqrt{5}$.
Therefore, $\cos \angle A H B=-\cos C=-\frac{\sqrt{6}}{6}$.