Connect MC and extend to point D on the circle, connect PD, and then connect PD⊥AB.
Reason: Because point M and point N are symmetrical about AB, when PD⊥AB, point P and point D are also symmetrical about AB.
Therefore, the line NP and MD must intersect at one point, and this point is on the axis of symmetry AB.
So after connecting NP and AB to find this point, we can find the D point where P is symmetrical about AB.
(2) As shown in the figure, connect AM and NO, extend NO to point C on the circle, and connect BC.
BC intersects AM at point D, connects DO and extends the intersection of both ends at points E and F, and then EF⊥AB.
Reason: Because when the diameter EF⊥ the diameter AB, point A and point B are symmetrical about EF,
Points C and M are symmetrical about EF, so the point where AM and BC intersect will be on EF.
And the symmetrical point c of point m about EF and the symmetrical point n of point m about ab will be on the same diameter.
So you can connect and expand NO to find point C.