(1) This problem is a typical tracing problem.

The distance between Party A and Party B is 90m, and the speed difference is 72/min-65/min=7/min, so the meeting time is 90/7.

At this time, B walked 72x90÷7=6480/7m, which is equivalent to walking 6480/7÷90, which is about 10.3 side.

10.3÷4 = 2 and 2.3, that is, B has walked for 2 weeks and has 2.3 sides, that is, A and B are on the side of ad.

So, the first question is D.

(2) Because the speed difference of 7/min is not a divisor of 90m, it is impossible for them to meet at the vertices of a square.