答案：
(1) 解：如答图，连接$OC$。$\because \angle AOB = 90^{\circ}$，点$C$是$AB$的中点，
$\therefore OC = AC = BC = \frac{1}{2}AB$，$\because$点$C$的横坐标为$-2$，$CE \perp x$轴于点$E$，
$\therefore E(-2,0)$，$CE$垂直平分$OA$，$\therefore AE = OE = 2$，
$\therefore OA = AE + OE = 2 + 2 = 4$，$\therefore A(-4,0)$。
$\because$直线$y = \frac{1}{2}x + b$经过点$A$，
$\therefore \frac{1}{2}×(-4) + b = 0$，解得$b = 2$。
(2) 证明：$\because AC = BC$，$CF \perp AB$，$\therefore CF$垂直平分$AB$，
$\therefore AF = BF$，$\therefore \angle ABF = \angle BAF$，
$\because BD \perp y$轴，$\therefore BD // x$轴，
$\therefore \angle ABD = \angle BAF$，
$\therefore \angle ABF = \angle ABD$，$\therefore$射线$BA$平分$\angle DBF$。
(3) 解：由(1)得$b = 2$，$OE = 2$，$OA = 4$，$\therefore y = \frac{1}{2}x + 2$，
当$x = 0$时，$y = 2$，$\therefore B(0,2)$，$\therefore OB = 2$，$\because OB^{2} + OF^{2} = BF^{2}$，且$BF = AF = 4 - OF$，$\therefore 2^{2} + OF^{2} = (4 - OF)^{2}$，解得$OF = \frac{3}{2}$，$\therefore EF = OE - OF = 2 - \frac{3}{2} = \frac{1}{2}$，
$\therefore EF$的长是$\frac{1}{2}$。