8 For the cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, the centers of faces $A B C D$ and $A D D^{\prime} A^{\prime}$ are points $M$ and $N$, respectively. The size of the angle formed by the skew lines $M N$ and $B D^{\prime}$ is $\qquad$.
8 Let the edge length of the cube be 1, and establish a spatial rectangular coordinate system with $\overrightarrow{D A}, \overrightarrow{D C}, \overrightarrow{D D^{\prime}}$ as the basis, as shown in the figure. Then we have $B(1,1,0), D^{\prime}(0,0,1)$, $M\left(\frac{1}{2}, \frac{1}{2}, 0\right), N\left(\frac{1}{2}, 0, \frac{1}{2}\right)$, therefore
$$
\begin{array}{l}
\overrightarrow{B D^{\prime}}=(-1,-1,1), \\
\overrightarrow{M N}=\left(0,-\frac{1}{2}, \frac{1}{2}\right) .
\end{array}
$$

Since
$$
\overrightarrow{B D^{\prime}} \cdot \overrightarrow{M N}=1,\left|\overrightarrow{B D^{\prime}}\right|=\sqrt{3},|\overrightarrow{M N}|=\frac{\sqrt{2}}{2} \text {, }
$$

Therefore,
$$
\cos =\frac{\overrightarrow{B D^{\prime}} \cdot \overrightarrow{M N}}{\left|\overrightarrow{B D^{\prime}}\right| \cdot|\overrightarrow{M N}|}=\frac{\sqrt{6}}{3} .
$$

Thus, the angle formed by the skew lines $M N$ and $B D^{\prime}$ is $\arccos \frac{\sqrt{6}}{3}$.