答：
①一般形如x=ay²抛物线焦点坐标是(1/(4a),0),所以题目中焦点坐标是(p/2,0).
②FA向量与x轴正向夹角为60°,则FA的斜率k=Tan( 60°)=√3
直线方程是：y - 0 = √3 (x - p/2)
y = √3(x- p/2) ,代入抛物线方程,求A点坐标：
3(x -p/2)² = 2px
3x² -3px +3/4*p² = 2px
3x² -5px +3/4*p² = 0
x = [5p + - √((5p)² - 4 * 3 *3/4*p² )]/6
x = [5p + - √(25p² - 9p²)]/6
x =（5p + - 4p)/6
x = 3p/2,p/6
y²= 3p² ,p²/3
|OA|² = x²+y² = 9/4 p² + 3p² = 21p²/4
或
|OA|² = x²+y² = p²/36 + p²/3 = 13p²/36
|OA| = √21/2 p
或
|OA| = √13/6 p
---完---