To solve the inequality \(2|x - 2| > |3x + 1|\), we consider the critical points where each absolute value expression changes sign.

1. Solve \(2(x - 2) = 3x + 1\) and \(2(x - 2) = -(3x + 1)\).

2. For \(2(x - 2) = 3x + 1\):

\(2x - 4 = 3x + 1\)

\(-4 - 1 = 3x - 2x\)

\(-5 = x\)

3. For \(2(x - 2) = -(3x + 1)\):

\(2x - 4 = -3x - 1\)

\(2x + 3x = -1 + 4\)

\(5x = 3\)

\(x = \frac{3}{5}\)

4. The critical values are \(x = -5\) and \(x = \frac{3}{5}\).

5. Test intervals around the critical points to determine where the inequality holds:

- For \(x < -5\), choose \(x = -6\):

\(2|-6 - 2| = 16\) and \(|3(-6) + 1| = 17\), so \(16 \not> 17\).

- For \(-5 < x < \frac{3}{5}\), choose \(x = 0\):

\(2|0 - 2| = 4\) and \(|3(0) + 1| = 1\), so \(4 > 1\).

- For \(x > \frac{3}{5}\), choose \(x = 1\):

\(2|1 - 2| = 2\) and \(|3(1) + 1| = 4\), so \(2 \not> 4\).

6. The solution is \(-5 < x < \frac{3}{5}\).

To solve the equation \(2|x - 1| = 3|x|\), consider the cases for the absolute values.

Case 1: \(x \geq 1\)

In this case, \(|x - 1| = x - 1\) and \(|x| = x\).

The equation becomes \(2(x - 1) = 3x\).

Simplifying gives \(2x - 2 = 3x\).

Rearranging, we get \(-2 = x\), which is not valid for \(x \geq 1\).

Case 2: \(0 \leq x < 1\)

Here, \(|x - 1| = 1 - x\) and \(|x| = x\).

The equation becomes \(2(1 - x) = 3x\).

Simplifying gives \(2 - 2x = 3x\).

Rearranging, we get \(2 = 5x\).

Thus, \(x = \frac{2}{5}\), which is valid for \(0 \leq x < 1\).

Case 3: \(x < 0\)

In this case, \(|x - 1| = 1 - x\) and \(|x| = -x\).

The equation becomes \(2(1 - x) = 3(-x)\).

Simplifying gives \(2 - 2x = -3x\).

Rearranging, we get \(2 = -x\).

Thus, \(x = -2\), which is valid for \(x < 0\).

Therefore, the solutions are \(x = -2\) and \(x = \frac{2}{5}\).

To solve the inequality \(|2x - 5| > 3|2x + 1|\), we consider two cases based on the definition of absolute value.

Case 1: \(2x - 5 > 3(2x + 1)\)

Simplifying, we get:

\(2x - 5 > 6x + 3\)

\(-5 - 3 > 6x - 2x\)

\(-8 > 4x\)

\(x < -2\)

Case 2: \(2x - 5 < -3(2x + 1)\)

Simplifying, we get:

\(2x - 5 < -6x - 3\)

\(2x + 6x < -3 + 5\)

\(8x < 2\)

\(x < \frac{1}{4}\)

Combining both cases, the solution is \(-2 < x < \frac{1}{4}\).

To solve the inequality \(|x - 2| > 2x - 3\), we consider two cases based on the definition of absolute value.

Case 1: When \(x - 2 \geq 0\), i.e., \(x \geq 2\).

The inequality becomes \(x - 2 > 2x - 3\).

Simplifying, we get:

\(x - 2 > 2x - 3\)

\(-2 + 3 > 2x - x\)

\(1 > x\)

Since \(x \geq 2\) and \(1 > x\) cannot both be true, there is no solution in this case.

Case 2: When \(x - 2 < 0\), i.e., \(x < 2\).

The inequality becomes \(-(x - 2) > 2x - 3\).

Simplifying, we get:

\(-x + 2 > 2x - 3\)

\(2 + 3 > 2x + x\)

\(5 > 3x\)

\(x < \frac{5}{3}\)

Since \(x < 2\) and \(x < \frac{5}{3}\) are consistent, the solution is \(x < \frac{5}{3}\).

To solve the inequality \(|3x - 1| < |2x + 5|\), we consider the cases for the absolute values.

Case 1: Both expressions are positive or both are negative.

1. Solve \(3x - 1 = 2x + 5\):

\(3x - 2x = 5 + 1\)

\(x = 6\)

2. Solve \(3x - 1 = -(2x + 5)\):

\(3x - 1 = -2x - 5\)

\(3x + 2x = -5 + 1\)

\(5x = -4\)

\(x = -\frac{4}{5}\)

Thus, the solution is \(-\frac{4}{5} < x < 6\).