5. In trapezoid $A B C D$, $A B / / C D$, the base angles $\angle D A B=36^{\circ}, \angle C B A=54^{\circ}, M$ and $N$ are the midpoints of sides $A B$ and $C D$, respectively. If the lower base $A B$ is exactly 2008 units longer than the upper base $C D$, then the line segment $M N=$ $\qquad$
5.1004.

As shown in Figure 4, draw $N S / / A D$ and $N T / / C B$ through point $N$. 
Then, from $\square A S N D$ and $\square B T N C$, we get
$D N=A S, N C=T B$,

and
$$
\begin{array}{l}
\angle N S T=\angle D A B=36^{\circ}, \\
\angle N T S=\angle C B A=54^{\circ} . \\
\text { Hence, } \angle S N T=180^{\circ}-(\angle N S T+\angle N T S) \\
=180^{\circ}-\left(36^{\circ}+54^{\circ}\right)=90^{\circ}
\end{array}
$$

We can determine that $\triangle N S T$ is a right triangle.
Noting that $A S=D N=N C=T B$, we get
$$
\begin{array}{l}
S T=A B-(A S+T B) \\
=A B-(D N+N C) \\
=A B-D C=2008 .
\end{array}
$$

Therefore, $M N=\frac{1}{2} S T=1004$.