Question1:where does △ def move, that is, what is the length of AD, and the connecting line between F and C is parallel to AB?

When the connecting line of FC is parallel to AB, the angle FCD=30 degrees, CD= root 3*FD=4 root 3cm,

So AD=AC-CD=2BC-CD= 12-4 root 3(cm).

Question 2: Where does △ def move, that is, what is the length of AD? Is the triangle with the length of line segments AD, FC and BC as three sides a right triangle?

Let the length of AD be x, then CD= 12-X, fc 2 = CD 2+FD 2 = (12-x) 2+4 2 =160-24x+x 2.

BC^2=6^2=36

A triangle with three sides AD, FC and BC is a right triangle. If FC is the hypotenuse, there are:

160-24x+x^2=X^2+36

AD=X=3 1/6cm

If AD is the hypotenuse, there are:

X^2= 160-24x+x^2+36

AD=X=49/6

If BC is the hypotenuse, there are:

36=X^2+ 160-24x+x^2

X 2- 12x+62 = 0 does not meet the requirements, so this situation does not exist.

Question ③: Is there a position to make ∠ fcd = 15 during the movement of △DEF?

∠FCD= 15

Tan∠fcd = 15 = 2- root 3

CD = FD/Tan∠FCD = 15 = 4/(2- root 3)=8+4 root 3>8+4 *1.7 =14.8 > 12=AC, so it is impossible, so there is no definite position, so that ∠ fcd = 15.