13. 4.3 * Given that $A$ is any point on the ellipse $x^{2}+4 y^{2}=4$, and $B$ is any point on the circle $x^{2}+(y-2)^{2}=\frac{1}{3}$. Find the maximum and minimum values of $|A B|$.
Assuming point $A$ is given, then the line connecting $A$ and point $M(0,2)$ intersects the circle at $B_{1}$ and $B_{2}$. Taking any point $B^{\prime}$ on the circle $M$, we have
$$
\begin{aligned}
\left|A B_{1}\right| & =|A M|-\left|B_{1} M\right|=|A M|-\left|B^{\prime} M\right| \\
& \leqslant\left|A B^{\prime}\right| \leqslant|A M|+\left|B^{\prime} M\right| \\
& =|A M|+\left|B_{2} M\right|=\left|A B_{2}\right| .
\end{aligned}
$$

Therefore, the point $B_{1}$ on the circle $M$ is the closest to $A$, and $B_{2}$ is the farthest from $A$. Thus, the problem is transformed into finding the maximum and minimum distances from a point on the ellipse to point $M$. Let the coordinates of point $A$ be $(x, y)$, then $|M A|^{2}=x^{2}+(y-2)^{2}$. Given $x^{2}+4 y^{2}=4$, we have
$$
|M A|^{2}=4-4 y^{2}+(y-2)^{2}=-3 y^{2}-4 y+8, \quad -1 \leqslant y \leqslant 1 .
$$

Thus, $|M A|_{\max }=\frac{2 \sqrt{21}}{3},|M A|_{\min }=1$. Therefore, $|A B|_{\max }=\frac{2 \sqrt{21}}{3}+\frac{\sqrt{3}}{3},|A B|_{\min }=1-\frac{\sqrt{3}}{3}$.