1.2. Given a cube with an edge of 1. Find the angle and distance between the skew diagonals of two of its adjacent faces.
1.2. Consider the diagonals $A B_{1}$ and $B D$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Since $B_{1} D_{1} \| B D$, the angle between the diagonals $A B_{1}$ and $B D$ is equal to the angle $A B_{1} D_{1}$. But the triangle $A B_{1} D_{1}$ is equilateral, so $\angle A B_{1} D_{1}=60^{\circ}$.

It is easy to check that the line $B D$ is perpendicular to the plane $A C A_{1} C_{1}$; therefore, when projected onto this plane, it passes through the midpoint $M$ of the segment $A C$. Similarly, the point $B_{1}$ is projected onto the midpoint $N$ of the segment $A_{1} C_{1}$. Thus, the distance between the lines $A B_{1}$ and $B D$ is equal to the distance from the point $A M$ to the line $A N$. If the legs of a right triangle are $a$ and $b$, and its hypotenuse is $c$, then the distance from the vertex of the right angle to the hypotenuse is $a b / c$. In the right triangle $A M N$, the legs are 1 and $1 / \sqrt{2}$, so its hypotenuse is $\sqrt{3 / 2}$, and the desired distance is $1 / \sqrt{ } \sqrt{3}$.