7. Let the ellipse $C_{1}: \frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ intersect the parabola $C_{2}$ : $y^{2}=8 x$ at a point $P\left(x_{0}, y_{0}\right)$. Define
$$
f(x)=\left\{\begin{array}{ll}
2 \sqrt{2 x} ; & 0<x \leq x_{0} \\
-2 \sqrt{2 x} ; & x>x_{0} .
\end{array}\right.
$$

If the line $y=a$ intersects $y=f(x)$ at points $A$ and $B$, and given the fixed point $N(2,0)$, then the range of the perimeter of $\triangle A B N$ is $\qquad$
7. $\left(\frac{20}{3}, 8\right)$.

From the problem, it is easy to get $x_{0}=\frac{4}{3}$.
Since $N(2,0)$ is exactly the common focus of the ellipse $C_{1}$ and the parabola $C_{2}$, by the definition of a parabola, we have
$$
|N A|=x_{A}+2 \text {. }
$$

Since point $B$ is on the ellipse, then
$$
\frac{x_{B}^{2}}{16}+\frac{y_{B}^{2}}{12}=1 \Rightarrow y_{B}^{2}=12\left(1-\frac{x_{B}^{2}}{16}\right) \text {. }
$$

Therefore, $|N B|=\sqrt{\left(x_{B}-2\right)^{2}+y_{B}^{2}}$
$$
\begin{array}{l}
=\sqrt{\left(x_{B}-2\right)^{2}+12\left(1-\frac{x_{B}^{2}}{16}\right)} \\
=\frac{1}{2} \sqrt{x_{B}^{2}-16 x_{B}+64}=4-\frac{1}{2} x_{B} .
\end{array}
$$

Since $|A B|=\left|x_{A}-x_{B}\right|=x_{B}-x_{A}$, and
$$
\frac{4}{3}=x_{0}<x_{B}<4 \text {, }
$$

Therefore, the perimeter of $\triangle A B N$ is
$$
\begin{array}{l}
x_{A}+2+4-\frac{1}{2} x_{B}+x_{B}-x_{A} \\
=6+\frac{1}{2} x_{B} \in\left(\frac{20}{3}, 8\right) .
\end{array}
$$