$$
\begin{array}{l}
\text { II. (16 points) Given } \\
\cos (\alpha+\beta)=\cos \alpha+\cos \beta .
\end{array}
$$

Find the maximum value of $\cos \alpha$.
From the problem, we have $\cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta = \cos \alpha + \cos \beta$.
Then, $(\cos \alpha - 1) \cos \beta - \sin \alpha \cdot \sin \beta - \cos \alpha = 0$.
Let point $P(\cos \beta, \sin \beta)$,
line $l: (\cos \alpha - 1) x - \sin \alpha y - \cos \alpha = 0$.
Then the trajectory equation of point $P$ is the unit circle $x^{2} + y^{2} = 1$, and point $P$ lies on line $l$.
Therefore, the distance from the center $O(0,0)$ to line $l$ is
$$
\begin{array}{l}
d = \frac{|-\cos \alpha|}{\sqrt{(\cos \alpha - 1)^{2} + (-\sin \alpha)^{2}}} \leqslant 1 \\
\Rightarrow \cos ^{2} \alpha + 2 \cos \alpha - 2 \leqslant 0 \\
\Rightarrow -1 \leqslant \cos \alpha \leqslant \sqrt{3} - 1 .
\end{array}
$$

Thus, the maximum value of $\cos \alpha$ is $\sqrt{3} - 1$.