To solve the inequality \(3|x - 1| < |2x + 1|\), we consider the non-modular inequality \((3(x - 1))^2 < (2x + 1)^2\).

Expanding both sides, we have:

\((3(x - 1))^2 = 9(x^2 - 2x + 1) = 9x^2 - 18x + 9\)

\((2x + 1)^2 = 4x^2 + 4x + 1\)

Setting the inequality:

\(9x^2 - 18x + 9 < 4x^2 + 4x + 1\)

Simplifying, we get:

\(5x^2 - 22x + 8 < 0\)

Solving the quadratic inequality \(5x^2 - 22x + 8 = 0\) gives the critical points:

\(x = \frac{2}{5}\) and \(x = 4\)

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

\(\frac{2}{5} < x < 4\)

To solve \(|x| < |5 + 2x|\), consider the critical points where \(x = 0\) and \(5 + 2x = 0\).

First, solve \(5 + 2x = 0\):

\(2x = -5\)

\(x = -\frac{5}{2}\)

Now, consider the intervals determined by the critical points \(x = 0\) and \(x = -\frac{5}{2}\).

1. For \(x < -5\), both \(x\) and \(5 + 2x\) are negative, so \(|x| = -x\) and \(|5 + 2x| = -(5 + 2x)\).

2. For \(-5 < x < -\frac{5}{3}\), \(x\) is negative and \(5 + 2x\) is positive, so \(|x| = -x\) and \(|5 + 2x| = 5 + 2x\).

3. For \(x > -\frac{5}{3}\), both \(x\) and \(5 + 2x\) are positive, so \(|x| = x\) and \(|5 + 2x| = 5 + 2x\).

Check each interval:

- For \(x < -5\), \(-x < -(5 + 2x)\) simplifies to \(x < -5\).

- For \(-5 < x < -\frac{5}{3}\), \(-x < 5 + 2x\) simplifies to \(x > -\frac{5}{3}\).

- For \(x > -\frac{5}{3}\), \(x < 5 + 2x\) simplifies to \(x < -5\), which is not possible.

Thus, the solution is \(x < -5\) or \(x > -\frac{5}{3}\).

To solve the inequality \(2|x - 3| > |3x + 1|\), we consider the critical points where the expressions inside the absolute values are zero.

1. Solve \(x - 3 = 0\) to get \(x = 3\).

2. Solve \(3x + 1 = 0\) to get \(x = -\frac{1}{3}\).

These critical points divide the number line into intervals: \((-\infty, -\frac{1}{3})\), \((-\frac{1}{3}, 3)\), and \((3, \infty)\).

Test each interval:

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

\(2|-1 - 3| = 2|4| = 8\) and \(|3(-1) + 1| = |-3 + 1| = 2\).

Since \(8 > 2\), this interval satisfies the inequality.

- For \(-\frac{1}{3} < x < 3\), choose \(x = 0\):

\(2|0 - 3| = 2|3| = 6\) and \(|3(0) + 1| = |1| = 1\).

Since \(6 > 1\), this interval satisfies the inequality.

- For \(x > 3\), choose \(x = 4\):

\(2|4 - 3| = 2|1| = 2\) and \(|3(4) + 1| = |12 + 1| = 13\).

Since \(2 < 13\), this interval does not satisfy the inequality.

Thus, the solution is \(-7 < x < 1\).

To solve \(|x - 3| > 2|x + 1|\), consider the critical points where each expression inside the absolute value is zero: \(x = 3\) and \(x = -1\).

Also consider where the expressions \(x - 3\) and \(x + 1\) change sign: \(x = 3\) and \(x = -1\).

Test intervals determined by these points: \((-\infty, -1)\), \((-1, 3)\), and \((3, \infty)\).

1. For \(x < -1\), both \(x - 3\) and \(x + 1\) are negative. The inequality becomes \(-(x - 3) > 2(-(x + 1))\), simplifying to \(x - 3 > 2x + 2\), which gives \(-3 > x + 2\) or \(-5 > x\). This is not possible in this interval.

2. For \(-1 < x < 3\), \(x - 3\) is negative and \(x + 1\) is positive. The inequality becomes \(-(x - 3) > 2(x + 1)\), simplifying to \(x - 3 < -2x - 2\), which gives \(3x < 1\) or \(x < \frac{1}{3}\).

3. For \(x > 3\), both \(x - 3\) and \(x + 1\) are positive. The inequality becomes \(x - 3 > 2(x + 1)\), simplifying to \(x - 3 > 2x + 2\), which gives \(-3 > x + 2\) or \(-5 > x\). This is not possible in this interval.

Combining the valid intervals, the solution is \(-5 < x < \frac{1}{3}\).

To solve the inequality \(|x + 3a| > 2|x - 2a|\), we consider the non-modular inequality \((x + 3a)^2 > (2(x - 2a))^2\).

Expanding both sides, we have:

\((x + 3a)^2 = x^2 + 6ax + 9a^2\)

\((2(x - 2a))^2 = 4(x^2 - 4ax + 4a^2) = 4x^2 - 16ax + 16a^2\)

Thus, the inequality becomes:

\(x^2 + 6ax + 9a^2 > 4x^2 - 16ax + 16a^2\)

Rearranging terms gives:

\(-3x^2 + 22ax - 7a^2 > 0\)

Solving the quadratic equation \(-3x^2 + 22ax - 7a^2 = 0\), we find the critical values:

\(x = \frac{1}{3}a\) and \(x = 7a\).

Therefore, the solution to the inequality is:

\(\frac{1}{3}a < x < 7a\).