One. (20 points) As shown in Figure 3, given that point $C$ is the midpoint of the minor arc $\overparen{A B}$ of $\odot O$, point $D$ is on $\overparen{A C}$, and $A C=2, A D + B D = \sqrt{6} + \sqrt{2}$. Find the degree measure of $\angle D A C$.
Obviously, $B C=A C$.
As shown in Figure 7, with $C$ as the center and $C A$ as the radius, draw $\odot C$, which must pass through point $B$. Extend $B C$ and $B D$ to intersect $\odot C$ at points $E$ and $F$, respectively, and connect $A E$, $A F$, $E F$, and $C F$. Then
$$
\begin{array}{l}
\angle D A C=\angle D B C \\
=\angle F B E=\angle E A F=\frac{1}{2} \angle E C F, \\
\angle E F B=90^{\circ}, C F=C A=2 .
\end{array}
$$

Therefore, $\angle F A D=\angle E A C$.
Also, $\angle E A C=\angle A E C=\angle A F B$, so $\angle F A D=\angle A F B \Rightarrow D F=A D$.
Since $B E=2 A C=4$, by the Pythagorean theorem, we get
$$
E F=\sqrt{B E^{2}-B F^{2}}=\sqrt{6}-\sqrt{2} \text {. }
$$

Thus, $S_{\triangle B F E}=\frac{1}{2} B F \cdot E F=2$.
Draw $F H \perp B E$ at point $H$, then
$$
F H=\frac{2 S_{\triangle B F E}}{B E}=1 \text {. }
$$

Therefore, in the right triangle $\triangle C H F$, we have $\sin \angle E C F=\frac{F H}{C F}=\frac{1}{2}$.

Hence, $\angle E C F=30^{\circ}$.
Therefore, $\angle D A C=\frac{1}{2} \angle E C F=15^{\circ}$.