)As shown in Fig. 1, in the quadrilateral pyramid $S-ABCD$, the side length of the square base $ABCD$ is $a$, and the side edges $SA=SB=SC=SD=2a$, $M$ is the midpoint of edge $SA$, and $N$ is the midpoint of edge $SC$. Find the cosine value of the angle formed by the skew lines $DM$ and $BN$.
Three, as shown in Figure 1, extend $B N$ to $P$ such that $B N = N P$, and extend $D M$ to $Q$ such that $D M = M Q$. Then $B C P S$ and $A D S Q$ are both parallelograms.
$\therefore S P \| B C$ II $A D \xlongequal{\|} Q S$.
Therefore, $Q$, $S$, and $P$ are collinear. $P Q = 2 B C = 2 A D$.
Since $A B = B C = C D = D A = a$, and $S A = S B = S C = S D = 2 a$, we have $P Q = 2 a$.

Draw a line through $P$ parallel to $Q I$ intersecting the extension of $A D$ at $D_{1}$, then $Q P D_{1} D$ is a parallelogram.
$$
P D_{1} = Q D, \quad D D_{1} = P Q = 2 a.
$$

Thus, $A U_{1} = 3 a$.
It is easy to see that $\angle B P D$ is the angle formed by the skew lines $B N$ and $D M$.
In $\square B C P S$, $S C^{2} + B P^{2} = 2 B C^{2} + 2 B S^{2}$,

which means $4 a^{2} + B P^{2} = 2 a^{2} + 8 a^{2}$.
Thus, $B P^{2} = 6 a^{2} \Rightarrow B P = \sqrt{6} a$.
Connect $B D_{1}$.
$$
\begin{array}{l}
B D_{1}^{2} = A B^{2} + A D_{1}^{2} = a^{2} + (3 a)^{2} = 10 a^{2}, \\
P D_{1} = Q D = B P = \sqrt{6} a.
\end{array}
$$

In $\triangle B P D_{1}$, applying the cosine rule, we get
$$
\begin{aligned}
\cos \angle B P D_{1} & = \frac{B P^{2} + P D_{1}^{2} - B D_{1}^{2}}{2 \cdot B P \cdot P D_{1}} \\
& = \frac{6 a^{2} + 6 a^{2} - 10 a^{2}}{2 \times \sqrt{6} a \times \sqrt{6} a} = \frac{1}{6}.
\end{aligned}
$$

Answer: The cosine value of the angle formed by the skew lines $D M$ and $B N$ is
$$
\frac{1}{6}
$$