Solution: Given the conditions in the topic, we set the ordinate of point P as Y. 。

Then: d 2 = (x-3) 2+y 2

=(5-3/5*x)^2.

De: 16x 2/25+y 2-25 = 0。 (@)

(1) Let the abscissa of point A be x 1, then it is represented graphically, x1>; 0, so when y=0 in (@), x 1=5.

So: AF=2, (1) is correct.

(2) Let the ordinate of point B be y2, then there is a graph, y2 >；; 0, so when x=0 in (@), y2=4.

So: BF=5, (2) is correct.

(3) by (1), (3) correct.

(4) Errors caused by (2) and (4).

To sum up: the correct one is (1)(2)(3).