Six, (1) The 5 shaded parts in Figure 4 can form a small pentagon.
$$\begin{array}{c} \because B F=A G=A H=E I=E K=D L=D M \\ =C N=C O=B P=4, \\ \angle B F B^{\prime}=\angle A G A^{\prime}=90^{\circ}, \\ \angle C O C^{\prime}=\angle B P B^{\prime}=90^{\circ}, \\ \angle D M D^{\prime}=\angle C N C^{\prime}=90^{\circ}, \\ \angle E K E^{\prime}=\angle D L D^{\prime}=90^{\circ}, \\ \angle A H A^{\prime}=\angle E I E^{\prime}=90^{\circ}, \\ \text { and ( } \left.\angle A^{\prime}+\angle B^{\prime}+\angle C^{\prime}+\angle D^{\prime}+\angle E^{\prime}\right)+(\angle 1 \\ +\angle 2+\angle 3+\angle 4+\angle 5)=5 \times 180^{\circ}, 91 j \\ \angle A^{\prime}+\angle B^{\prime}+\angle C^{\prime}+\angle D^{\prime}+\angle E^{\prime} \\ =(\angle-2) \times 180^{\circ}=3 \times 180^{\circ} . \\ \therefore \angle 1+\angle 2+\angle 3+\angle 4+\angle 5=360^{\circ} . \end{array}$$

The 5 shaded parts in Figure 4 can form a small pentagon as shown in Figure 6 (enlarged).
(2) The additional perimeter of the polygon is equal to
Figure 6
$A^{\prime} H+A^{\prime} G+B^{\prime} F+$
$B^{\prime} P+C^{\prime} O+C^{\prime} N+D^{\prime} M+D^{\prime} L+E^{\prime} K+E^{\prime} I$.
This value is exactly the perimeter of the pentagon formed by the shaded areas in Figure 4, and it is evident that this pentagon has an inscribed circle with a radius of 4.

Since the perimeter $s$ of the pentagon is greater than the circumference of its inscribed circle, the circumference of the inscribed circle is $8 \pi, \pi>3.14$,
then $s>8 \pi>8 \times 3.14=25.12>25$.
Therefore, the perimeter of the new pentagon increases by at least 25 units.
(Provided by Tang Wenqing, Hainan Middle School, Haimen City, Jiangsu Province)