10 As shown in the figure, in the triangular prism $A B C-A_{1} B_{1} C_{1}$, the lateral edge $A A_{1} \perp$ plane $A B C$, triangle $A B C$ is an equilateral triangle with a side length of 2, $M$ is a point on $A A_{1}$, $A A_{1}=4, A_{1} M=$ 1. $P$ is a point on the edge $B C$ such that the shortest path from $P$ along the lateral surface of the prism, passing through the edge $C C_{1}$, to point $M$ is $3 \sqrt{2}$. Let the intersection point of this shortest path with $C C_{1}$ be $N$.
(1) Prove that $A_{1} B / /$ plane $M N P$;
(2) Find the tangent value of the dihedral angle (acute) formed by plane $M N P$ and plane $A B C$.
10 (1) Since the lateral edge $A A_{1} \perp$ plane $A B C$, and triangle $A B C$ is an equilateral triangle, therefore, all the lateral faces are congruent rectangles. As shown in the figure, rotate the lateral face $B C_{1}$ by $120^{\circ}$ so that it lies in the same plane as the lateral face $A C_{1}$, and point $P$ moves to the position of point $P_{1}$. Connect $M P_{1}$, then $M P_{1}$ is the shortest path from point $P$ along the lateral edges of the prism through $C C_{1}$ to point $M$.
Let $P C=x$, then $P_{1} C=x$.
In the right triangle $\triangle M A P_{1}$, $(2+x)^{2}+3^{2}=18$, so $x=1$, thus point $P$ is the midpoint of $B C$. It is easy to see that $N C=1$. Let $A_{1} C$ intersect $M N$ at point $Q$, then point $Q$ is the midpoint of $A_{1} C$, so $A_{1} B / / P Q$, so $A_{1} B / /$ plane $M N P$.
(2) Connect $P P_{1}$, then $P P_{1}$ is the intersection line of plane $M N P$ and plane $A B C$, draw $N H \perp$ $P P_{1}$ at $H$, and since $C C_{1} \perp$ plane $A B C$, connect $C H$, it is easy to see that $C H \perp P P_{1}$.

Therefore, $\angle N H C$ is the plane angle of the dihedral angle formed by plane $N M P$ and plane $A B C$.
In the right triangle $\triangle P H C$, since $\angle P C H=\frac{1}{2} \angle P C P_{1}=60^{\circ}$, so $C H=\frac{1}{2}$. In the right triangle $\triangle N C H$, $\tan \angle N H C=\frac{N C}{C H}=2$.