在 $\frac{1}{(2+ax)^5}$ 的展开式中，$x^3$ 的系数是 $\sqrt{4-ax}$ 展开式中 $x^2$ 系数的 5 倍，其中 $a$ 是非零常数。求 $a$ 的确切值。

1. Write $\frac{1}{(2+ax)^5}$ in the standard form $k(1+bx)^n$.

$\frac{1}{(2+ax)^5} = (2+ax)^{-5} = \left[2\left(1+\frac{a}{2}x\right)\right]^{-5} = \frac{1}{32}\left(1+\frac{a}{2}x\right)^{-5}$

2. Evaluate the coefficient of $x^3$ in the expansion of $(2+ax)^{-5}$.

The term in $x^3$ is:
$\frac{1}{32} \left[ \frac{(-5)(-6)(-7)}{3!} \left(\frac{a}{2}x\right)^3 \right]$
$= \frac{1}{32} (-35) \left(\frac{a^3}{8}x^3\right) = -\frac{35a^3}{256} x^3$
So, the coefficient of $x^3$ is $-\frac{35a^3}{256}$.

3. Write $\sqrt{4-ax}$ in the standard form $k(1+bx)^n$.

$\sqrt{4-ax} = (4-ax)^{\frac{1}{2}} = \left[4\left(1-\frac{a}{4}x\right)\right]^{\frac{1}{2}} = 2\left(1-\frac{a}{4}x\right)^{\frac{1}{2}}$

4. Evaluate the coefficient of $x^2$ in the expansion of $\sqrt{4-ax}$.

The term in $x^2$ is:
$2 \left[ \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!} \left(-\frac{a}{4}x\right)^2 \right]$
$= 2 \left(-\frac{1}{8}\right) \left(\frac{a^2}{16}x^2\right) = -\frac{a^2}{64} x^2$
So, the coefficient of $x^2$ is $-\frac{a^2}{64}$.

5. Form the equation based on the given relationship and solve for $a$.

We are given that the coefficient of $x^3$ is 5 times the coefficient of $x^2$:
$-\frac{35a^3}{256} = 5 \left( -\frac{a^2}{64} \right)$
$-\frac{35a^3}{256} = -\frac{5a^2}{64}$
Since $a$ is a non-zero constant ($a \neq 0$), we can divide both sides by $-5a^2$:
$\frac{7a}{256} = \frac{1}{64}$
$7a = \frac{256}{64}$
$7a = 4 \implies a = \frac{4}{7}$