If the function $$f(x) = \begin{cases} \frac{\sqrt{2+\cos x}-1}{(\pi-x)^2}, & x \neq \pi \\ k, & x = \pi \end{cases}$$ is continuous at $$x = \pi$$, then k equals:

To ensure the function $$f(x)$$ is continuous at $$x = \pi$$, the limit as $$x$$ approaches $$\pi$$ must equal the value of the function at $$x = \pi$$, which is $$k$$. Therefore, we need to evaluate:

$$\lim_{x \to \pi} \frac{\sqrt{2+\cos x}-1}{(\pi-x)^2} = k$$

As $$x$$ approaches $$\pi$$, the numerator $$\sqrt{2+\cos x} - 1$$ approaches $$\sqrt{2 + \cos \pi} - 1 = \sqrt{2 + (-1)} - 1 = \sqrt{1} - 1 = 0$$, and the denominator $$(\pi - x)^2$$ approaches $$0$$. This results in a $$\frac{0}{0}$$ indeterminate form, so we can use L'Hôpital's rule or a substitution to simplify.

Substitute $$t = \pi - x$$. As $$x \to \pi$$, $$t \to 0$$. Also, $$\cos x = \cos(\pi - t) = -\cos t$$. The expression becomes:

$$\lim_{t \to 0} \frac{\sqrt{2 - \cos t} - 1}{t^2}$$

At $$t = 0$$, the numerator is $$\sqrt{2 - \cos 0} - 1 = \sqrt{2 - 1} - 1 = \sqrt{1} - 1 = 0$$ and the denominator is $$0$$, confirming the $$\frac{0}{0}$$ form. Apply L'Hôpital's rule by differentiating the numerator and denominator with respect to $$t$$.

Let $$f(t) = \sqrt{2 - \cos t} - 1$$ and $$g(t) = t^2$$.

The derivative of $$f(t)$$ is:

$$f'(t) = \frac{d}{dt} \left( (2 - \cos t)^{1/2} - 1 \right) = \frac{1}{2} (2 - \cos t)^{-1/2} \cdot \sin t = \frac{\sin t}{2 \sqrt{2 - \cos t}}$$

The derivative of $$g(t)$$ is $$g'(t) = 2t$$.

By L'Hôpital's rule:

$$\lim_{t \to 0} \frac{f(t)}{g(t)} = \lim_{t \to 0} \frac{f'(t)}{g'(t)} = \lim_{t \to 0} \frac{\frac{\sin t}{2 \sqrt{2 - \cos t}}}{2t} = \lim_{t \to 0} \frac{\sin t}{4t \sqrt{2 - \cos t}}$$

This limit can be separated as:

$$\frac{1}{4} \lim_{t \to 0} \frac{\sin t}{t} \cdot \lim_{t \to 0} \frac{1}{\sqrt{2 - \cos t}}$$

We know $$\lim_{t \to 0} \frac{\sin t}{t} = 1$$. For the second limit, as $$t \to 0$$, $$\cos t \to 1$$, so $$2 - \cos t \to 1$$, and $$\sqrt{2 - \cos t} \to 1$$. Thus:

$$\lim_{t \to 0} \frac{1}{\sqrt{2 - \cos t}} = \frac{1}{\sqrt{1}} = 1$$

Therefore, the limit is:

$$\frac{1}{4} \cdot 1 \cdot 1 = \frac{1}{4}$$

So, $$\lim_{x \to \pi} f(x) = \frac{1}{4}$$. For continuity at $$x = \pi$$, $$f(\pi) = k$$ must equal this limit:

$$k = \frac{1}{4}$$

Comparing with the options:

A. $$\frac{1}{4}$$

B. $$0$$

C. $$2$$

D. $$\frac{1}{2}$$

Hence, the correct answer is Option A.