Solve : `(|x-1|-3)(|x+2)-5) lt 0`.

Explanation

To solve the inequality \((|x-1|-3)(|x+2|-5) < 0\), we will analyze the expression step by step.

### Step 1: Identify critical points
The critical points occur where the expressions inside the absolute values equal zero or where the entire expression equals zero.

1. For \(|x-1| = 3\):
- \(x - 1 = 3 \Rightarrow x = 4\)
- \(x - 1 = -3 \Rightarrow x = -2\)

2. For \(|x+2| = 5\):
- \(x + 2 = 5 \Rightarrow x = 3\)
- \(x + 2 = -5 \Rightarrow x = -7\)

The critical points are \(x = -7, -2, 3, 4\).

### Step 2: Test intervals
We will test the intervals defined by these critical points:
1. \((- \infty, -7)\)
2. \((-7, -2)\)
3. \((-2, 3)\)
4. \((3, 4)\)
5. \((4, \infty)\)

### Step 3: Evaluate the sign of the expression in each interval

1. **Interval \((- \infty, -7)\)**:
- Choose \(x = -8\):
\[ (|-8-1|-3)(|-8+2|-5) = (|-9|-3)(|-6|-5) = (9-3)(6-5) = 6 > 0 \]

2. **Interval \((-7, -2)\)**:
- Choose \(x = -5\):
\[ (|-5-1|-3)(|-5+2|-5) = (|-6|-3)(|-3|-5) = (6-3)(3-5) = 3 \cdot (-2) = -6 < 0 \]

3. **Interval \((-2, 3)\)**:
- Choose \(x = 0\):
\[ (|0-1|-3)(|0+2|-5) = (|-1|-3)(|2|-5) = (1-3)(2-5) = (-2)(-3) = 6 > 0 \]

4. **Interval \((3, 4)\)**:
- Choose \(x = 3.5\):
\[ (|3.5-1|-3)(|3.5+2|-5) = (|2.5|-3)(|5.5|-5) = (2.5-3)(5.5-5) = (-0.5)(0.5) = -0.25 < 0 \]

5. **Interval \((4, \infty)\)**:
- Choose \(x = 5\):
\[ (|5-1|-3)(|5+2|-5) = (|4|-3)(|7|-5) = (4-3)(7-5) = 1 \cdot 2 = 2 > 0 \]

### Step 4: Combine results
The expression is negative in the intervals \((-7, -2)\) and \((3, 4)\).

### Final Solution
Thus, the solution to the inequality \((|x-1|-3)(|x+2|-5) < 0\) is:
\[ x \in (-7, -2) \cup (3, 4) \]