(1) The parametric equation of curve C is $$\begin{cases} \overset{x=4+2cos\beta }{y=2sin\beta }\end{cases}$$ (β is the parameter, β∈$$0，π$$ ).

Converting to rectangular coordinate equation: (x-4)2+y2\=4 (y≥0).

The parametric equation of line l is $$\begin{cases} \overset{x=tcos\alpha }{y=tsin\alpha }\end{cases}$$ (t is the parameter, α is the inclination angle).

Converting to polar coordinate equation: θ=α.

(2) From (1), we know that curve C is a semicircle.

If line l and curve C have exactly one common point P, then line l is tangent to the semicircle.

Let P be (ρ, θ). From the problem, we know: $$sinθ= \frac {1}{2}$$,

So: $$θ= \frac {π}{6}$$,

So: ρ2+22\=42,

Solving for ρ: $$ρ=2 \sqrt {3}$$.

Therefore: Point P is ($$\boxed{2 \sqrt {3}, \frac {π}{6}}$$).