The first question has been answered very well upstairs.

2.？

As shown in the figure, triangle AFC and triangle ABE are folded along AF and AE respectively. Branch 1 ∠ 1 = ∠ 2 = 45.

Therefore ∠ fge = 90. So the triangle is a right triangle.

(If strict proof is required, it is as follows: Fold the triangular AFC along the AF into a triangular AFG. Connect GE.

It is proved that all triangles AEB are equal to triangles AEG. Then ∠ 1 = ∠ 2 = 45

Therefore ∠ fge = 90. Direct evidence.

3.

Rotate triangle ABC to triangle AEF as shown in the figure.

Judging from the meaning of the question, DC=DF=BC+EF= 1. DA=DA。

AC=AF again. Therefore, the triangular ADC is completely equal to the triangular ADF.

It doesn't look like it, but it is.

The area of triangle ADF is 1× 1 divided by 2= 1/2.

So the area of the hexagon is 1/2×2= 1.

Just a moment, please.