∴△ADE≌△CAB，
∴AD=CA=4,DE=AB=2，
∴OD=OA+AD= 1+4=5，
∴E point coordinates are (5,2);
∫A( 1，0)，B(3，0)，C( 1，4)，
∴AC=4，
∴D(5,0)，
∴ The vertical line of segment AD is x= 1+52=3, and the vertical line of segment AC is y=2.
∴q(3,2)；
(II) The existence of such points T (72 72,0) and (52,0) can make △PEF a right triangle.
There are two situations:
(i) Let EF⊥AE intersect the X axis at F when △PFE takes point E as the vertex of right angle, as shown in figure 1.
∵△AED∽△EFD，
∴DFDE=EDAD= 12，
∴DF= 12DE= 1，
∴ point f (6 6,0),
∴ point t (72 72,0);
(2) When △ p ′ f ′ e takes point f ′ as the vertex of right angle, as shown in the figure.
∫△AED∽△EF′D，
∴df′de=dead= 12，
∴df′= 12de= 1，
∴ point f' (4,0),
Point t (52 52,0).
To sum up (1) and (2), it is known that the T coordinates of the points that meet the conditions are (72,0) and (52,0).