5. Given $P$ as any point on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{15}=1$, and $EF$ as any diameter of the circle $\odot N:(x-1)^{2}+y^{2}=4$. Then the range of $\overrightarrow{P E} \cdot \overrightarrow{P F}$ is $\qquad$
5. $[5,21]$.

Let $N$ be the center, and
$$
\begin{array}{l}
\overrightarrow{P E}=\overrightarrow{P N}+\overrightarrow{N E}, \\
\overrightarrow{P F}=\overrightarrow{P N}+\overrightarrow{N F}=\overrightarrow{P N}-\overrightarrow{N E},
\end{array}
$$

we know
$$
\begin{array}{l}
\overrightarrow{P E} \cdot \overrightarrow{P F}=(\overrightarrow{P N}+\overrightarrow{N E})(\overrightarrow{P N}-\overrightarrow{N E}) \\
=|\overrightarrow{P N}|^{2}-|\overrightarrow{N E}|^{2}=|\overrightarrow{P N}|^{2}-4 .
\end{array}
$$

Since $N$ is exactly the right focus of the ellipse, the focal radius $|\overrightarrow{P N}| \in[a-c, a+c]$.
Given $a=4, c=1$, we have $|\overrightarrow{P N}| \in[3,5]$. Therefore, the required range is $[5,21]$.