(1) sets the coincidence after t seconds.

From the meaning of the question, it can be concluded that the m movement distance +AB=N movement distance should be made.

∴2t=t+ 12

The solution is t = 12.

∴ After points M and N move 12 seconds, points M and N coincide (in C).

(2) When an equilateral triangle is formed, AM=AN=MN, that is, △AMN is similar to △ABC.

Let t seconds later form an equilateral triangle.

Before n passes through a:

When △AMN∽△ACB:

By AM/AC=AN/AB

t/ 12 =( 12-2t)/ 12。

The solution is t=4.

N after a:

When m and n are on AC ∠ man = 0, the condition that △AMN is similar to △ABC is not satisfied.

After the two points M and N coincide with C, move again until it stops. Because the distance between m and n is always less than the length of CB, and ∠MAN is always less than 60, the condition that △AMN is similar to △ABC is not satisfied.

To sum up, after moving point M and point N for 4 seconds, an equilateral triangle △AMN can be obtained.

.

(3) existence.

Let t seconds later form an isosceles triangle.

Starting from points m and n moving to the edge of BC,

Then the isosceles triangle with MN as the base can only be △AMN, without △BMN and △CMN.

∫ Take MN as the base

∴AM=AN。

When m moves slower than n, when it can exist:

∠∠B =∠C，AB=AC，AM=AN

∴△ACM≌△ABN

∴CM=BN

From the coincidence of m and n in C after 12s:

1 *(t- 12)= 12-2 *(t- 12)

The solution is t= 16.

When point M and point N move on the edge of BC, an isosceles triangle with MN as the base can be obtained. At this time, the moving time of m and n is 16 seconds.