(2) connect OA. As can be seen from the coordinate values of A and B, OA is perpendicular to OB (but your drawing is a little inaccurate, and in your drawing, OA is obviously not perpendicular to OB), and AOB is a right angle. Crossing point a is parallel to AM, equal to OB, and connects BM. Because the internal angle AOB is a right angle, it is obvious that the parallelogram BOAM is a rectangle. It is not difficult to determine the coordinates (3, 4) of the point m.

(3) As long as it is proved that the BOE angle is a right angle. Because AB=AE, AC=AG, and angle EAC= angle BAG (that is, right-angle BAE+ angle BAC= right-angle GAC+ angle BAC, which is obviously true), all triangle bags are equal to triangle EAC. Therefore, angle CEA= angle ABG. Let AB and OE intersect at point K, then triangle KAE is similar to triangle KOB (because angle BKO= angle EKA, it has been proved that angle CEA= angle ABG), then angle EAB= angle BOE. Because the EAB angle is the inner corner of a square, it should be at right angles to BOE. Obviously, connecting BE, in the triangle BOE, Bo 2+OE 2 = Be 2, so the quadrilateral OBPE is a Pythagorean quadrilateral.