(I) Since the right focus of the ellipse C: $$\frac {x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}=l$$ (a>b>0) is at F(1,0) and its eccentricity is $$\frac {1}{2}$$, we have the following system of equations:
$$\begin{cases} c=1 \\ \frac {c}{a}= \frac {1}{2} \\ a^{2}=b^{2}+c^{2}\end{cases}$$
Solving this system, we get a=2 and b=$$\sqrt {3}$$.

Therefore, the equation of the ellipse C is $$\frac {x^{2}}{4}+ \frac {y^{2}}{3}=1$$.

(II) Let M be the point (x₁, y₁) and N be the point (x₂, y₂). From the system of equations:
$$\begin{cases} y=k(x-4) \\ \frac {x^{2}}{4}+ \frac {y^{2}}{3}=1\end{cases}$$
we obtain the quadratic equation (4k²+3)x²-32k²x+(64k²-12)=0.

According to the problem, the discriminant Δ=(−32k²)²−4(4k²+3)(64k²−12)>0, which leads to 0<k²<$$\frac {1}{4}$$.

Hence, $$x_{1}+x_{2}= \frac {32k^{2}}{4k^{2}+3}$$ and x₁x₂=$$\frac {64k^{2}-12}{4k^{2}+3}$$.

Note that when x₁=1 or x₂=1, we get k²=$$\frac {1}{4}$$, which contradicts the problem's condition.

Since k_MF + k_NF = $$\frac {y_{1}}{x_{1}-1}+ \frac {y_{2}}{x_{2}-1}$$ = $$\frac {k(x_{1}-4)}{x_{1}-1}$$ + $$\frac {k(x_{2}-4)}{x_{2}-1}$$ = $$\frac {k[2x_{1}x_{2}-5(x_{1}+x_{2})+8]}{(x_{1}-1)(x_{2}-1)}$$ = $$\frac {k[2( \frac {64k^{2}-12}{4k^{2}+3})-5( \frac {32k^{2}}{4k^{2}+3})+8]}{(x_{1}-1)(x_{2}-1)}$$ = 0,

we conclude that the angles formed by lines MF and NF with the x-axis are complementary, i.e., $\boxed{\text{the slopes of MF and NF are negative reciprocals of each other}}$.