Step 1: Write the continuity condition.
For $f$ to be continuous at $x=\pi$, we need $k=\displaystyle\lim_{x\to\pi}\dfrac{\sqrt{2+\cos x}-1}{(\pi-x)^2}$.
Step 2: Rationalise the top.
Multiply top and bottom by $\sqrt{2+\cos x}+1$: \[ \frac{(2+\cos x)-1}{(\pi-x)^2(\sqrt{2+\cos x}+1)}=\frac{1+\cos x}{(\pi-x)^2(\sqrt{2+\cos x}+1)}. \]
Step 3: Use a half-angle identity.
Since $1+\cos x=2\cos^2\dfrac{x}{2}$, the top becomes $2\cos^2\dfrac{x}{2}$.
Step 4: Rewrite the cosine near $x=\pi$.
Note $\cos\dfrac{x}{2}=\sin\left(\dfrac{\pi-x}{2}\right)$, so the top is $2\sin^2\left(\dfrac{\pi-x}{2}\right)$.
Step 5: Apply the small-angle limit.
As $x\to\pi$, $\sin\left(\dfrac{\pi-x}{2}\right)\approx\dfrac{\pi-x}{2}$, so $2\sin^2\left(\dfrac{\pi-x}{2}\right)\approx 2\cdot\dfrac{(\pi-x)^2}{4}=\dfrac{(\pi-x)^2}{2}$.
Step 6: Cancel and finish.
\[ \lim_{x\to\pi}\frac{(\pi-x)^2/2}{(\pi-x)^2(\sqrt{2+\cos x}+1)}=\frac{1}{2(1+1)}=\frac14. \] So $k=\dfrac14$. \[ \boxed{\frac14} \]