7. The function $y=2 \cos \left(x+\frac{\pi}{4}\right) \cos \left(x-\frac{\pi}{4}\right)+\sqrt{3} \sin 2 x$ has a maximum value and a minimum positive period. The sum of these two values is
7. $2+\pi$ To determine the properties of the given function, we first need to simplify it. For $2 \cos \left(x+\frac{\pi}{4}\right) \cos \left(x-\frac{\pi}{4}\right)$, it seems we should use the product-to-sum formula, but we can also simplify using the complementary relationship between $x+\frac{\pi}{4}$ and $\frac{\pi}{4}-x$ and the trigonometric identities.
By the trigonometric identities, we get $\cos \left(x-\frac{\pi}{4}\right)=\cos \left(\frac{\pi}{4}-x\right)=\sin \left[\frac{\pi}{2}-\left(\frac{\pi}{4}-x\right)\right]=\sin \left(\frac{\pi}{4}+x\right)$. Therefore,
$$
\begin{aligned}
y & =2 \cos \left(x+\frac{\pi}{4}\right) \sin \left(\frac{\pi}{4}+x\right)+\sqrt{3} \sin 2 x=\sin \left(\frac{\pi}{2}+2 x\right)+\sqrt{3} \sin 2 x \\
& =\sqrt{3} \sin 2 x+\cos 2 x=2 \sin \left(2 x+\frac{\pi}{6}\right) .
\end{aligned}
$$

Thus, the range of the function $y=\cos \left(x+\frac{\pi}{4}\right) \cos \left(x-\frac{\pi}{4}\right)+\sqrt{3} \sin 2 x$ is $[-2,2]$. The smallest positive period is $\pi$.