To solve the inequality \(2 - 3x < |x - 3|\), we consider the critical points where the expressions inside the absolute value change sign.

1. Consider the equation \(2 - 3x = x - 3\).

2. Solving for \(x\), we have:

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

\(2 + 3 = x + 3x\)

\(5 = 4x\)

\(x = -\frac{1}{2}\)

3. The critical value is \(x = -\frac{1}{2}\).

4. Test intervals around the critical point:

- For \(x > -\frac{1}{2}\), choose \(x = 0\):

\(2 - 3(0) < |0 - 3|\)

\(2 < 3\), which is true.

- For \(x < -\frac{1}{2}\), choose \(x = -1\):

\(2 - 3(-1) < |-1 - 3|\)

\(5 < 4\), which is false.

5. Therefore, the solution is \(x > -\frac{1}{2}\).

To solve the inequality \(|x - 2| > 3|2x + 1|\), we first consider the non-modular inequality:

\((x - 2)^2 > (3(2x + 1))^2\)

This simplifies to:

\((x - 2)^2 > 9(2x + 1)^2\)

Expanding both sides, we get:

\(x^2 - 4x + 4 > 36x^2 + 72x + 36\)

Rearranging terms gives:

\(x^2 - 4x + 4 - 36x^2 - 72x - 36 > 0\)

\(-35x^2 - 76x - 32 > 0\)

Solving the quadratic equation \(-35x^2 - 76x - 32 = 0\) gives critical values \(x = -1\) and \(x = -\frac{1}{7}\).

Testing intervals, we find the solution to the inequality is:

\(-1 < x < -\frac{1}{7}\)

To solve the inequality \(2x > |x - 1|\), we need to consider the definition of absolute value, which gives us two cases:

Case 1: \(x - 1 \\geq 0\), which implies \(x \\geq 1\).

In this case, \(|x - 1| = x - 1\), so the inequality becomes:

\(2x > x - 1\)

Solving for \(x\), we get:

\(2x - x > -1\)

\(x > -1\)

Since \(x \\geq 1\) is a condition for this case, the solution for this case is \(x \\geq 1\).

Case 2: \(x - 1 < 0\), which implies \(x < 1\).

In this case, \(|x - 1| = -(x - 1) = -x + 1\), so the inequality becomes:

\(2x > -x + 1\)

Solving for \(x\), we get:

\(2x + x > 1\)

\(3x > 1\)

\(x > \frac{1}{3}\)

Since \(x < 1\) is a condition for this case, the solution for this case is \(\frac{1}{3} < x < 1\).

Combining both cases, the overall solution is \(x > \frac{1}{3}\).

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

Case 1: \(2x + 1 \\geq 0\)

In this case, \(|2x + 1| = 2x + 1\).

So, the inequality becomes \(3x + 5 < 2x + 1\).

Subtract \(2x\) from both sides: \(x + 5 < 1\).

Subtract 5 from both sides: \(x < -4\).

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

In this case, \(|2x + 1| = -(2x + 1)\).

So, the inequality becomes \(3x + 5 < -(2x + 1)\).

Which simplifies to \(3x + 5 < -2x - 1\).

Add \(2x\) to both sides: \(5x + 5 < -1\).

Subtract 5 from both sides: \(5x < -6\).

Divide by 5: \(x < -\frac{6}{5}\).

The critical value is \(x = -\frac{6}{5}\), and the solution is \(x < -\frac{6}{5}\).

To solve the inequality \(|x - 3a| > |x - a|\), we consider the critical points where the expressions inside the absolute values are zero. These points are \(x = 3a\) and \(x = a\).

We analyze the inequality in the intervals determined by these critical points: \((-\infty, a)\), \((a, 3a)\), and \((3a, \infty)\).

1. For \(x < a\):

\(|x - 3a| = 3a - x\) and \(|x - a| = a - x\).

The inequality becomes \(3a - x > a - x\), which simplifies to \(3a > a\), always true.

2. For \(a < x < 3a\):

\(|x - 3a| = 3a - x\) and \(|x - a| = x - a\).

The inequality becomes \(3a - x > x - a\), which simplifies to \(2a > 2x\) or \(x < 2a\).

3. For \(x > 3a\):

\(|x - 3a| = x - 3a\) and \(|x - a| = x - a\).

The inequality becomes \(x - 3a > x - a\), which simplifies to \(-3a > -a\), never true.

Thus, the solution is \(x < 2a\).