一、(50 Points)
$M$ is the midpoint of side $BC$ of $\triangle ABC$, $\angle MAC=15^{\circ}$, find the maximum value of $\angle B$.
One, (50 points)
$\angle B$ maximum value is $105^{\circ}$
Draw line segment $B C$, take midpoint $M$, and construct a circular segment with $MC$ as a chord, such that the central angle of the arc $MC$ is $30^{\circ}$. Then the major arc is the locus of point $A$. Draw a tangent from point $B$ to this arc, and take the tangent point as $A$, so that $\angle B$ achieves its maximum value.

From $B A^{2}=B M \cdot B C$, we get $\triangle A B M \backsim \Omega$ $\triangle C B A$, with the similarity ratio being $A B: B C=1: \sqrt{2}$.

Construct an isosceles right triangle $A C O$ with $A C$ as the hypotenuse, then
$$
O A=O C=\frac{\sqrt{2}}{2} A C=A M,
$$

Since $\angle O A M=\angle O A C+\angle C A M=45^{\circ}+15^{\circ}=60^{\circ}$, $\triangle A M O$ is an equilateral triangle,
so $O M=O A=O C$,
so $O$ is the circumcenter of $\triangle A M C$.
Since $B A$ is a tangent to $O$,
so $\angle B A O=90^{\circ}$,
so $\angle B A M=30^{\circ}=\angle B C A$,
so $\angle B=180^{\circ}-\angle B C A-\angle C A B=105^{\circ}$.