详细解答上面问题的解题过程。---**Question Number and Points:** 27. (8分) **Problem Description:** 已知射线OM平分∠AOB, 点C为OM上任意一点，过点C作直线l//OB交射线OA于点D. **Part (1):** (1) 如图1, 若∠OCD = 30°, 则∠AOB = ______ °. **Figure 1 Description:** * Type: Geometric diagram. * Elements: * Point O: Origin. * Rays OA, OB, OM: Emanating from O. OM is between OA and OB. * Line l: Passes through C, parallel to OB. * Point C: On ray OM. * Point D: Intersection of line l and ray OA. * Labels: A, B, C, D, O, M, l. * Angle indicated: ∠OCD. * Text Label: 图1 (Figure 1). **Part (2):** (2) 点E是射线DC上一动点 (不与点C, D重合), OF平分∠DOE交CD于点F, 过点F作FG//OM交OA于点G. **Figure 2 Description:** * Type: Geometric diagram. * Elements: * Point O: Origin. * Rays OA, OB, OM: Emanating from O. OM is between OA and OB. * Line segment CD: Part of the line l from Figure 1. E is on the extension of CD or DC. (Based on the text, E is on ray DC). * Points C, D, E: Collinear. E is on ray DC but not C or D. * Ray OF: Emanating from O, bisects ∠DOE, intersects CD at F. * Point F: On line segment CD. * Line segment FG: Starts at F, parallel to OM, intersects OA at G. * Point G: On ray OA. * Labels: A, B, C, D, E, F, G, O, M, l. * Text Label: 图2 (Figure 2). **Part (2)①:** ① 如图2, 若∠OCD = 60°, 当OE⊥CD时, 求∠OFG的度数. **Part (2)②:** ② 当点E在运动过程中, 设∠OFG = α, ∠OEC = β, 直接写出α和β之间的数量关系.