9. Let the function $f(x)=a x^{2}+8 x+3(a<0)$, for a given negative number $a$, there is a maximum positive number $l(a)$ such that the inequality $|f(x)| \leqslant 5$ holds for the entire interval $\left[0, l(a)^{\prime}\right]$. Question: For what value of $a$ is $l(a)$ maximized? Find this maximum $l(a)$.

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The function $f(x) = ax^2 + 8x + 3$ is a quadratic function with a negative leading coefficient $a$, meaning it opens downwards. We need to find the maximum positive number $l(a)$ such that $|f(x)| \leq 5$ for all $x$ in the interval $[0, l(a)]$.

First, we need to determine the points where $f(x) = 5$ and $f(x) = -5$. These points will help us find the interval where $|f(x)| \leq 5$.

1. Solve $f(x) = 5$:
   \[
   ax^2 + 8x + 3 = 5 \implies ax^2 + 8x - 2 = 0
   \]
   Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get:
   \[
   x = \frac{-8 \pm \sqrt{64 + 8a}}{2a} = \frac{-8 \pm \sqrt{64 + 8a}}{2a}
   \]
   Let the roots be $x_1$ and $x_2$.

2. Solve $f(x) = -5$:
   \[
   ax^2 + 8x + 3 = -5 \implies ax^2 + 8x + 8 = 0
   \]
   Using the quadratic formula again, we get:
   \[
   x = \frac{-8 \pm \sqrt{64 - 32a}}{2a} = \frac{-8 \pm \sqrt{64 - 32a}}{2a}
   \]
   Let the roots be $x_3$ and $x_4$.

The interval where $|f(x)| \leq 5$ is between the smallest and largest roots of these equations. We need to find the value of $a$ that maximizes the length of this interval.

The length of the interval is given by the difference between the largest and smallest roots. For the interval to be maximized, the roots should be as far apart as possible. This occurs when the discriminants of the quadratic equations are maximized.

For the equation $ax^2 + 8x - 2 = 0$, the discriminant is:
\[
\Delta_1 = 64 + 8a
\]
For the equation $ax^2 + 8x + 8 = 0$, the discriminant is:
\[
\Delta_2 = 64 - 32a
\]

To maximize the interval, we need to balance these discriminants. The maximum interval length occurs when the roots of the two equations are symmetric around the vertex of the parabola. The vertex of the parabola $f(x) = ax^2 + 8x + 3$ is at:
\[
x = -\frac{8}{2a} = -\frac{4}{a}
\]

For the interval to be symmetric around the vertex, the roots of the equations should be equidistant from the vertex. This occurs when the discriminants are equal:
\[
64 + 8a = 64 - 32a \implies 40a = 0 \implies a = -2
\]

Substituting $a = -2$ into the equations, we get:
\[
x_1, x_2 = \frac{-8 \pm \sqrt{64 - 16}}{-4} = \frac{-8 \pm \sqrt{48}}{-4} = \frac{-8 \pm 4\sqrt{3}}{-4} = 2 \mp \sqrt{3}
\]
\[
x_3, x_4 = \frac{-8 \pm \sqrt{64 + 64}}{-4} = \frac{-8 \pm \sqrt{128}}{-4} = \frac{-8 \pm 8\sqrt{2}}{-4} = 2 \mp 2\sqrt{2}
\]

The interval where $|f(x)| \leq 5$ is between the smallest and largest roots:
\[
l(a) = (2 + \sqrt{3}) - (2 - 2\sqrt{2}) = \sqrt{3} + 2\sqrt{2}
\]

Thus, the value of $a$ that maximizes $l(a)$ is $a = -2$, and the maximum $l(a)$ is:
\[
\boxed{\sqrt{3} + 2\sqrt{2}}
\]
9. $f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}, f_{\max }=3-\frac{16}{a}$.
(1) $3-\frac{16}{a}>5$, i.e., $-8-\frac{4}{a}$, so $l(a)$ is the larger root of the equation $a x^{2}+8 x+3=-5$, $l(a)=\frac{-8-\sqrt{64-32 a}}{2 a}=\frac{4}{\sqrt{4-2 a}-2} \leqslant \frac{4}{\sqrt{20}-2}=\frac{\sqrt{5}+1}{2}$, equality holds only when $a=-8$. Also, $\frac{\sqrt{5}+1}{2}>\frac{1}{2}$, therefore, $l(a)$ has a maximum value of $\frac{\sqrt{5}+1}{2}$ if and only if $a=-8$.