5. In isosceles $\triangle A B C$, it is known that $A C=B C=\sqrt{5}$, points $D, E, F$ are on sides $A B, B C, C A$ respectively, and $A D=D B=E F=1$. If $\overrightarrow{D E} \cdot \overrightarrow{D F} \leqslant \frac{25}{16}$, then the range of $\overrightarrow{E F} \cdot \overrightarrow{B A}$ is
5. $\left[\frac{4}{3}, 2\right]$.

Let $D$ be the origin, and the rays $D B$ and $D C$ be the positive directions of the $x$ and $y$ axes, respectively, in a Cartesian coordinate system. Then
$$
A(-1,0), B(1,0), C(0,2) \text {. }
$$

Let the points $E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$, where,
$$
y_{1}=-2 x_{1}+2, y_{2}=2 x_{2}+2 \text {. }
$$

Let the midpoint of segment $E F$ be $M\left(x_{0}, y_{0}\right)$. Then
$$
\left\{\begin{array}{l}
y_{1}-y_{2}=-2\left(x_{1}+x_{2}\right)=-4 x_{0}, \\
x_{1}-x_{2}=2-\frac{y_{1}+y_{2}}{2}=2-y_{0} .
\end{array}\right.
$$

Given $E F=1$, we have
$$
\left(-4 x_{0}\right)^{2}+\left(2-y_{0}\right)^{2}=1 \text {. }
$$
$$
\begin{array}{l}
\text { Hence }\left(-4 x_{0}\right)^{2}=1-\left(2-y_{0}\right)^{2} \geqslant 0 \\
\Rightarrow 1 \leqslant y_{0} \leqslant 3 . \\
\text { Also, } \overrightarrow{D E} \cdot \overrightarrow{D F}=\frac{1}{4}\left[(\overrightarrow{D E}+\overrightarrow{D F})^{2}-(\overrightarrow{D E}-\overrightarrow{D F})^{2}\right] \\
=\overrightarrow{D M}^{2}-\overrightarrow{E F}^{2}=\overrightarrow{D M}^{2}-\frac{1}{4} \leqslant \frac{25}{16} \\
\Rightarrow \overrightarrow{D M}^{2} \leqslant \frac{29}{16} \\
\Rightarrow x_{0}^{2}+y_{0}^{2} \leqslant \frac{29}{16} .
\end{array}
$$

Substituting equation (1) into equation (3), and eliminating $x_{0}$, we get
$$
\begin{array}{l}
15 y_{0}^{2}-4 y_{0}^{2}-32 \leqslant 0 \\
\Rightarrow-\frac{8}{5} \leqslant y_{0} \leqslant \frac{4}{3} .
\end{array}
$$

Combining equations (2) and (4), we have $1 \leqslant y_{0} \leqslant \frac{4}{3}$. Thus, $\frac{2}{3} \leqslant x_{1}-x_{2} \leqslant 1$.
Therefore, $\overrightarrow{E F} \cdot \overrightarrow{B A}=\left(x_{2}-x_{1}, y_{2}-y_{1}\right) \cdot(-2,0)$
$$
=2\left(x_{1}-x_{2}\right) \in\left[\frac{4}{3}, 2\right] \text {. }
$$