5. A polygon is cut with one cut (the cut does not pass through the vertices of the polygon) to divide it into 2 polygons, then one of the polygons is cut with another cut (the cut does not pass through the vertices of the polygon) to divide it into another polygon, ... and so on. If starting from a square, to cut out a triangle, a quadrilateral, a pentagon, ... up to a 2008-sided polygon, the minimum number of cuts required is
5.2011014.

Let a total of $k$ cuts be made. Since each cut increases the number of polygons by one, there are a total of $k+1$ polygons. Apart from one triangle, one quadrilateral, one pentagon, ... one 2008-sided polygon, there are
$$
(k+1)-2006=k-2005
$$

polygons.
Considering the total sum $S$ of the number of sides of these polygons, then
$$
\begin{array}{l}
S \geqslant 3+4+\cdots+2008+3(k-2005) \\
=3 k+2011018 .
\end{array}
$$

Since each cut adds 4 sides (the original polygon has two sides split into two, and the cut line adds two new sides), thus, $S=4+4 k$.
Therefore, $4+4 k=S \geqslant 3 k+2011018$, which means
$$
k \geqslant 2011014 \text {. }
$$

On the other hand, first cut the square into a triangle and a pentagon with one cut, then cut the pentagon into a quadrilateral and a pentagon with another cut, and again cut the pentagon into a quadrilateral and a pentagon, then cut the quadrilateral into a triangle and a pentagon, and finally cut the pentagon into a triangle and a hexagon. Up to this point, a total of five cuts have been made, resulting in two triangles, one quadrilateral, one pentagon, and one hexagon. Now, operate on one of the triangles, each cut producing a new triangle, and four cuts produce a heptagon and several triangles. Thereafter, always choose a triangle and cut $t-3$ times to produce a $t(t=8,9, \cdots, 2008)$-sided polygon and several triangles. In this way, a total of $5+4+5+\cdots+2005=2011014$ cuts have been made.
Therefore, the minimum value of $k$ is 2011014.