That is, when t=3 seconds, the quadrilateral APQD is a parallelogram;

(2) (1) When the point P moves from M to B, if P, E, D, E and D are on the same straight line, make the PH in H vertical to AD:

∴ah=bp=bm-pm=4-t,hd=ad-ah=2+t； And ph = ba = 3 √ 3; ∠PDH=∠EPQ=60？ .

tan∠PDH=tan60？ =PH/HD=(3√3)/(2+t)，t = 1；

(2) When the point P moves from B to M, if P, E, D, E and D are on the same straight line, make the PH vertical to AD at H:

Then ah = BP = t-4, HD = ad-ah = 6-(t-4) =10-t.

tan∠PDH=tan60？ =(3√3)/(t-4)，t=7。

Therefore, when t= 1 or 7 seconds, P, E and D are three-point * * * lines;

(3)① When point P moves to point B, PB= 1, then PE=PQ=PM+MQ=6, EM=3√3=BA.

∴S overlapping part = s ⊿ pqe = pq * em/2 = 6 * (3 √ 3)/2 = 9 √ 3;

② When point P moves from point B to point M, PB= 1, then t=5=MQ and PQ=8=PE.

Let PE cross AD at F, then AF=BM=4 and FD=AD-AF=2.

S overlap = (FD+PC) * AB/2 = (2+7) * (3 √ 3)/2 = (27 √ 3)/2.

(4) When point P coincides with point B, that is, t=3 seconds, the AD part covered by △EPQ reaches the maximum value;

When point P returns from point B, BP= 1, that is, t=5 seconds, EQ just passes point D, and the maximum value ends.

Therefore, when t satisfies 3 seconds ≤t≤5 seconds, the covered line segment keeps the maximum value.