(1)∵点A是圆心∴AE=AC,∴∠C=∠AEC,
∵AN⊥EC,∴∠C+∠CAN=90°,
∵EO⊥BM,∴∠AEC+∠EMO=90°,
∴∠CAN=∠EMO,
又∵∠CAN=∠CBN+∠ANB,∠EBO=∠MBN+∠MNB
∠CBN=∠MBN
∴∠ANB=∠MNB,
∵BC是直径∴∠BEC=90°,∵AN⊥CE,∴AN‖BE,∴∠ANB=∠NBE,
∴∠MNB=∠NBE,∴BE=NE;
(2)由(1)知AN‖BE,∵点A是BC的中点,∴点N是CE的中点,∴tanC=BE/CE=1/2
(3)∵直线y=kx+3与y轴交于点A,∴OA=3,∴BO^2+3^2=r^2
∵AN‖BE,∴S△ABE=S△NBE,∴BO×AE=BE×EN,即BOr=BE^2,
∵CE=2BE,∴5BE^2=BC^2,即BE^2=0.8r^2
∴BO=0.8r,∴(0.8r)^2+9=r^2,r=5