4. As shown in Figure 2, given that $\odot O$ is the incircle of rhombus $A B C D$ with side length 16, points $E$ and $F$ are on sides $A B$ and $B C$ respectively, and $E F$ is tangent to $\odot O$ at point $M$. If $B E=4, B F=13$, then the length of $E F$ is
4. 10. 5 .

As shown in Figure 6, let
$\odot O$ be tangent to $AB$ at
point $N$, and connect $AC$,
$BD$, $OE$, $OF$,
$ON$.
From the given conditions, we have
$$
\begin{aligned}
& \angle COF \\
= & 180^{\circ}-\angle OFC-\angle OCF \\
= & 90^{\circ}-\frac{1}{2} \angle EFC+\frac{1}{2} \angle EBF \\
= & \frac{1}{2}(\angle BFE+\angle EBF)=\angle AEO .
\end{aligned}
$$

Thus, $\triangle AOE \sim \triangle CFO$.
Then $\frac{OA}{CF}=\frac{AE}{OC} \Rightarrow \frac{OA}{16-13}=\frac{16-4}{OA}$
$$
\Rightarrow OA=6 \text {. }
$$

In the right triangle $\triangle OAB$, by the projection theorem, we get
$$
AN=\frac{OA^2}{AB}=\frac{36}{16}=\frac{9}{4} \text {. }
$$

Then $BN=AB-AN=16-\frac{9}{4}=\frac{55}{4}$.
Therefore, $EF=2BN-BE-BF=10.5$.