Expand R (x - lambda) using the compound angle formula.

We have:
$R \cos \lambda = 1 \quad (1)$
$R \sin \lambda = \sqrt{3} \quad (2)$

Square and add equations (1) and (2):

Using the identity $\cos^2 \lambda + \sin^2 \lambda = 1$:

Since $R > 0$, $R = \sqrt{4} = 2$.

Divide equation (2) by equation (1):

Given $0 < \lambda < \frac{\pi}{2}$, $\lambda$ is in the first quadrant.

The expression for $f(x)$ is $\boxed{2 \cos \left(x - \frac{\pi}{3}\right)}$.

From part (a), $f(x) = 2 \cos \left(x - \frac{\pi}{3}\right)$.
The range of $\cos \theta$ is $[-1, 1]$.
Therefore, the range of $f(x)$ is $2 \times [-1, 1] = [-2, 2]$.

Since $f(x)$ ranges from $-2$ to 2, the maximum value of $|f(x)|$ is 2. This occurs when $f(x) = 2$ or $f(x) = -2$.

To minimize the fraction $\frac{1}{1+|f(x)|}$, its denominator $1+|f(x)|$ must be maximized.
The maximum value of $|f(x)|$ is 2.
So, the maximum value of $1+|f(x)|$ is $1+2 = 3$.
The minimum value of $\frac{1}{1+|f(x)|}$ is $\frac{1}{3}$.
The minimum value is $\boxed{\frac{1}{3}}$.

Using the expression for $f(x)$ from part (a):

Let $Y = x - \frac{\pi}{3}$. We need to solve $\cos Y = \frac{\sqrt{3}}{2}$.
The principal value for $Y$ is $\arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$.

The general solution for $\cos Y = \cos \alpha$ is $Y = 2n\pi \pm \alpha$, where $n \in \mathbb{Z}$.
So, $x - \frac{\pi}{3} = 2n\pi \pm \frac{\pi}{6}$.

Case 1: $x - \frac{\pi}{3} = 2n\pi + \frac{\pi}{6}$

Case 2: $x - \frac{\pi}{3} = 2n\pi - \frac{\pi}{6}$

The general solution is $\boxed{x = 2n\pi + \frac{\pi}{2} \text{ or } x = 2n\pi + \frac{\pi}{6}, \text{ where } n \in \mathbb{Z}}$.