∵ quadrilateral ABCD is a rectangle,

∴AM∥DN.

∴∠KNM=∠ 1.

∵∠ 1=70 ,

∴∠KNM=∠KMN=∠ 1=70，

∴∠MKN=40。

So the answer is: 40;

(2) isosceles,

Reason: ∫ab∨CD, ∴∠ 1=∠MND.

∫ origami along MN, ∴∠ 1=∠kmn∠MND =∠kmn,

∴km=kn；

So the answer is: isosceles;

(3) As shown in Figure 2, when the minimum area of △KMN is 12, KN=BC= 1, so KN⊥B'M,

∠∠NMB =∠KMN∠KMB = 90，

∴∠ 1 =∠ NMB = 45。 Similarly, when the paper is folded down, ∠ 1 =∠ NMB = 135.

So the answer is: 45 or 135 (just write one); ?

(4) There are two situations:

Situation 1: As shown in Figure 3, fold the rectangular piece of paper in half so that point B and point D coincide, and point K also coincides with point D. 。

MK=MB=x, then am = 5-X.

According to Pythagorean Theorem, 12+(5-x)2=x2,

Solution x = 2.6.

∴MD=ND=2.6.

S△MNK = S△MND = 12× 1×2.6 = 1.3。

Case 2: As shown in Figure 4, the rectangular paper is folded in half along the diagonal AC, and the crease is AC.

MK=AK=CK=x, then DK = 5-X.

Similarly, MK = NK = 2.6 can be obtained.

∫MD = 1，

∴s△mnk= 12× 1×2.6= 1.3.

The maximum area of △MNK is 1.3.