1. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $O$ is the center of face $A B C D$, and $O_{1}$ is the center of face $A D D_{1} A_{1}$. Find the cosine value of the angle formed by the skew lines $D_{1} O$ and $B O_{1}$.
1. Connect $A D_{1}, A C$, let $E$ be the midpoint of $O A$, then $O_{1} E / / D_{1} O$, so $\angle E O_{1} B$ is the angle formed by $D_{1} O$ and $B O_{1}$, and $O_{1} E=\frac{1}{2} D_{1} O$. Suppose the edge length of the cube is 1, then $B O_{1}=D_{1} 0=\sqrt{A B^{2}+A O_{1}^{2}}=\frac{\sqrt{6}}{2}, O_{1} E=\frac{\sqrt{6}}{4}, B E=$ $\sqrt{O E^{2}+B O^{2}}=\frac{\sqrt{10}}{4}$. In $\triangle B O_{1} E$, $\cos \angle B O_{1} E=\frac{5}{6}$ is the answer.