6. If the circle $x^{2}+y^{2}=k^{2}$ covers at least one maximum point and one minimum point of the function $f(x)=\sqrt{3} \sin \frac{\pi x}{k}$, find the range of values for $k$.

The function $f(x)=\sqrt{3} \sin \frac{\pi x}{k}$ has a maximum value of $\sqrt{3}$ and a minimum value of $-\sqrt{3}$. The points where these maximum and minimum values occur are given by:
- Maximum points: $(x, \sqrt{3})$ where $x = k \left(2n + \frac{1}{2}\right)$ for integers $n$.
- Minimum points: $(x, -\sqrt{3})$ where $x = k \left(2n - \frac{1}{2}\right)$ for integers $n$.

For the circle $x^{2}+y^{2}=k^{2}$ to cover at least one maximum point and one minimum point, the points $(x, \sqrt{3})$ and $(x, -\sqrt{3})$ must lie within or on the boundary of the circle. This means that the distance from the origin to these points must be less than or equal to $k$.

The distance from the origin to the point $(x, \sqrt{3})$ is:
\[
\sqrt{x^2 + (\sqrt{3})^2} = \sqrt{x^2 + 3}
\]
For the point to be within or on the circle, we need:
\[
\sqrt{x^2 + 3} \leq k
\]
Squaring both sides, we get:
\[
x^2 + 3 \leq k^2
\]
Since $x = k \left(2n + \frac{1}{2}\right)$, we substitute this into the inequality:
\[
\left(k \left(2n + \frac{1}{2}\right)\right)^2 + 3 \leq k^2
\]
Simplifying, we get:
\[
k^2 \left(2n + \frac{1}{2}\right)^2 + 3 \leq k^2
\]
\[
k^2 \left(4n^2 + 2n + \frac{1}{4}\right) + 3 \leq k^2
\]
\[
4n^2 k^2 + 2n k^2 + \frac{1}{4} k^2 + 3 \leq k^2
\]
\[
4n^2 k^2 + 2n k^2 + \frac{1}{4} k^2 \leq k^2 - 3
\]
\[
4n^2 k^2 + 2n k^2 + \frac{1}{4} k^2 \leq k^2 - 3
\]
\[
4n^2 k^2 + 2n k^2 + \frac{1}{4} k^2 - k^2 \leq -3
\]
\[
4n^2 k^2 + 2n k^2 - \frac{3}{4} k^2 \leq -3
\]
\[
k^2 \left(4n^2 + 2n - \frac{3}{4}\right) \leq -3
\]

For the inequality to hold, the term inside the parentheses must be non-positive. However, since $k^2$ is always positive, the term inside the parentheses must be non-positive. This implies that the smallest value of $k$ must be such that the inequality holds for the smallest $n$ (i.e., $n=0$).

For $n=0$:
\[
k^2 \left(0 + 0 - \frac{3}{4}\right) \leq -3
\]
\[
-\frac{3}{4} k^2 \leq -3
\]
\[
k^2 \geq 4
\]
\[
k \geq 2
\]

Thus, the range of values for $k$ is:
\[
\boxed{k \geq 2}
\]
6. Since $f(x)=\sqrt{3} \sin \frac{n \pi}{k}$ is an odd function, its graph is symmetric about the origin, so the circle $x^{2}+y^{2}=k^{2}$ only needs to cover one of the extreme points of $f(x)$.

Let $\frac{n \pi}{k}=\frac{\pi}{2}$, solving for the nearest maximum point to the origin $P\left(\frac{k}{2}, \sqrt{3}\right)$, according to the problem, we get $k^{2} \geqslant\left(\frac{k}{2}\right)^{2}+(\sqrt{3})^{2}$, solving this yields $|k| \geqslant 2$.