1. Given the quadratic function
$$
f(x)=a x^{2}+\left(16-a^{3}\right) x-16 a^{2}(a>0)
$$

the graph of which intersects the $x$-axis at points $A$ and $B$. Then the minimum length of segment $A B$ is $\qquad$.
-1.12 .
The factorization yields $f(x)=\left(x-a^{2}\right)(a x+16)$. Thus, $A\left(a^{2}, 0\right), B\left(-\frac{16}{a}, 0\right)$.
Therefore, the length of segment $A B$ is $a^{2}+\frac{16}{a}$. Let $F(a)=a^{2}+\frac{16}{a}$, then
$$
F^{\prime}(a)=2 a-\frac{16}{a^{2}}=\frac{2\left(a^{3}-8\right)}{a^{2}} .
$$

Hence, $F(a)$ is monotonically decreasing on the interval $(0,2)$ and monotonically increasing on the interval $(2,+\infty)$.
Therefore, the minimum value of $F(a)$ is
$$
F(2)=2^{2}+\frac{16}{2}=12 \text {. }
$$