Only one. The midline of the line segment OM passing through the midpoint of OM intersects with AD at point P, and the coordinate of point P is (0.5,4) (principle: the distance from any point of the midline of the line segment to both ends of the line segment is equal).

So is there a possible point P on AB? First of all, if the point P is on AB, the angle POM is obtuse, and the PM side is the longest, which can only be OP=OM, but in fact OP is the hypotenuse of the right triangle PBO, that is, the distance from the point P on the AB side to the far point O, that is, OP >；; OB & gt3.5, therefore, there is no qualified point p on the AB side;

So will there be a qualified point p next to the CD? First of all, if there is a point P, the angle PMO is obtuse and the side PO is the longest, which can only be PM=OM. Actually CM=8. Since the angle MCP is a right angle, the length of PM on the hypotenuse must be greater than CM, so PM cannot be equal to OM.

The second question:

P has four (2,4) (0,4) (4,4) (-7/2, the root of two is fifteen).

Process: Point 1: Make the vertical line of OM, pass AD at point P, and connect OP and PM, then OP=PM, and the coordinate of point P is (2,4) (principle: the distance from any point on the vertical line of a line segment to both ends of the line segment is equal).

The second point: Let AD intersect with Y axis at point P, then OP=4, followed by PM. Since OM=4 and OP=OM=4, then point P (0 0,4).

The third point: Make a straight line PM perpendicular to the X axis, intersect with AD at point P, and connect OP, then OM=MP=4, then point P (4,4).

Point 4: If there is a point P on AB, the coordinate is (-7/2, y). Since the triangle POM is an isosceles triangle, PO=OM=4. Then, the square of OB+the square of BP = the square of OP, that is, the square of 49/4 +y = 16, that is, y = the root number fifteen.

Then there is still doubt, will there be a qualified P point on the CD? Assuming that there is, then CM=5, then PM is the hypotenuse length of right triangle CMP >; 5, that is, PM> Hmm. So there will be no qualified P point next to the CD.

The third problem is that we make the middle vertical line of OM and intersect with AD at point P, and the coordinate of this P is (2.5,4).

Let the coordinates of point P be (x, 4) on AD, and the Y axis of AD is at point E..

Connect OP, because the triangle MOP is an isosceles triangle, and OP=MO=5.

Because OE=4, EP= absolute value X.

So, the square of OE+the square of EP = the square of OP.

That is, 16+x =25 squared, then x = plus or minus 3.

That is, the coordinate of point P is (-3,4) (3,4).

Then will there be a qualified P point next to AB? Suppose there is, the coordinate of point P is (-7/2, y), and OP=OM=5.

Because the square of BP+the square of OB = the square of OP, that is, the square of Y +49/4 =25, and the root sign of Y = 5 1

So, is there a qualified P point next to the CD? Assuming there is, it needs PM=OM=5. Let's see if it is possible: the coordinate of p is (9, y). First, because CM=4 and the square of CM+the square of CP = the square of PM, that is, the square of 16 +y =25, the solution is y = plus or minus 3.

First, the negative 3 is discarded, because the point P is on the CD, so the coordinate of the point P is (9,3), which holds.

So the answer to the third question is (2.5,4) (-3,4) (3,4) (-7/2, the root of 2 is 51) (9,3), and five.