已知 $x < 0$，解不等式 $\frac{2}{x} < |x| - a$，其中 $a$ 为常数且满足 $a > 3$。

1. Eliminate the modulus sign based on the given domain.

Since $x < 0$, we know that $|x| = -x$.
Substitute this into the given inequality:
$\frac{2}{x} < -x - a$

2. Convert the rational inequality into a quadratic inequality.

Multiply both sides by $x$. Because $x < 0$, we must reverse the inequality sign:
$2 > x(-x - a)$
$2 > -x^2 - ax$
Rearranging the terms yields a standard quadratic inequality:
$x^2 + ax + 2 > 0$

3. Find the critical values by solving the corresponding quadratic equation.

Set $x^2 + ax + 2 = 0$ and use the quadratic formula to find the critical values:
$x = \frac{-a \pm \sqrt{a^2 - 4(1)(2)}}{2} = \frac{-a \pm \sqrt{a^2 - 8}}{2}$
Since $a > 3$, we have $a^2 > 9$, which ensures $a^2 - 8 > 0$ and thus real, distinct roots.

4. Determine the solution to the quadratic inequality and intersect it with $x < 0$.

The parabola $y = x^2 + ax + 2$ opens upwards, so it is strictly positive outside its roots:
$x < \frac{-a - \sqrt{a^2 - 8}}{2} \quad \text{or} \quad x > \frac{-a + \sqrt{a^2 - 8}}{2}$
Since $a > 3$, we know $\sqrt{a^2 - 8} < \sqrt{a^2} = a$. This means the larger root is negative:
$\frac{-a + \sqrt{a^2 - 8}}{2} < 0$
Given the initial domain restriction $x < 0$, both regions must be bounded by $0$. Thus, the final solution is:
$x < \frac{-a - \sqrt{a^2 - 8}}{2} \quad \text{or} \quad \frac{-a + \sqrt{a^2 - 8}}{2} < x < 0$