(7) $\left(\frac{20}{3}, 8\right)$ Hint: Given that one of the intersection points of the ellipse $C_{1}: \frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ and the parabola $C_{2}: y^{2}=$ $8 x$ is $P\left(x_{0}, y_{0}\right)$, and $x_{0}>0$, we 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 know $|N A|=x_{A}+2$. Since point $B$ is on the ellipse, we have $\frac{x_{B}^{2}}{16}+\frac{y_{B}^{2}}{12}=1, y_{B}^{2}=12\left(1-\frac{x_{B}^{2}}{16}\right)$. Therefore,
$$
\begin{aligned}
|N B| & =\sqrt{\left(x_{B}-2\right)^{2}+y_{B}^{2}} \\
& =\sqrt{\left(x_{B}-2\right)^{2}+12\left(1-\frac{x_{B}^{2}}{16}\right)} \\
& =\frac{1}{2} \sqrt{x_{B}^{2}-16 x_{B}+64}=4-\frac{1}{2} x_{B} .
\end{aligned}
$$

Since $|A B|=\left|x_{A}-x_{B}\right|=x_{B}-x_{A}$, we have
$$
\text{The perimeter of } \triangle A B N =x_{A}+2+4-\frac{1}{2} x_{B}+x_{B}-x_{A}=6+\frac{1}{2} x_{B} \text{. }
$$

Since $\frac{4}{3}=x_{0}<x_{B}<4$, we have $\frac{20}{3}<6+\frac{1}{2} x_{B}<8$. Therefore, the range of the perimeter of $\triangle A B N$ is $\left(\frac{20}{3}, 8\right)$.
(7) Let the ellipse $C_{1}: \frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ and the parabola $C_{2}: y^{2}=8 x$ intersect at a point $P\left(x_{0}, y_{0}\right)$, and define
$$
f(x)=\left\{\begin{array}{ll}
2 \sqrt{2 x}, & 0<x \leq x_{0}, \\
\frac{12 \sqrt{2}}{\sqrt{x}}, & x>x_{0},
\end{array}\right.
$$

If the line $y=a$ intersects $f(x)$ 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