(a) The graph of \(y = |4x - 2|\) is a V-shaped graph. The vertex of the graph is at \(x = \frac{1}{2}\), \(y = 0\). The graph is symmetrical about the vertical line \(x = \frac{1}{2}\) and extends into the second quadrant.

(b) To solve \(1 + 3x < |4x - 2|\), consider two cases:

Case 1: \(4x - 2 \geq 0\) (i.e., \(x \geq \frac{1}{2}\))

Then \(|4x - 2| = 4x - 2\).

Solving \(1 + 3x < 4x - 2\) gives:

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

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

\(1 - x < -2\)

\(-x < -3\)

\(x > 3\)

Case 2: \(4x - 2 < 0\) (i.e., \(x < \frac{1}{2}\))

Then \(|4x - 2| = -(4x - 2) = -4x + 2\).

Solving \(1 + 3x < -4x + 2\) gives:

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

\(1 + 3x + 4x < 2\)

\(1 + 7x < 2\)

\(7x < 1\)

\(x < \frac{1}{2}\)

Thus, the solution is \(x < \frac{1}{2}\) or \(x > 3\).

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

\((3x - a)^2 > 2^2(x + 2a)^2\).

This simplifies to:

\((3x - a)^2 > 4(x + 2a)^2\).

Expanding both sides, we get:

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

Simplifying further:

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

Rearranging terms gives:

\(5x^2 - 22ax - 15a^2 > 0\).

We solve the quadratic equation \(5x^2 - 22ax - 15a^2 = 0\) to find critical values:

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 5\), \(b = -22a\), \(c = -15a^2\), we find:

\(x = \frac{22a \pm \sqrt{(-22a)^2 - 4 \cdot 5 \cdot (-15a^2)}}{2 \cdot 5}\).

\(x = \frac{22a \pm \sqrt{484a^2 + 300a^2}}{10}\).

\(x = \frac{22a \pm \sqrt{784a^2}}{10}\).

\(x = \frac{22a \pm 28a}{10}\).

This gives critical values \(x = 5a\) and \(x = -\frac{3}{5}a\).

The solution to the inequality is:

\(x > 5a\) or \(x < -\frac{3}{5}a\).

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

Expanding both sides, we get:

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

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

Setting up the inequality:

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

Rearranging terms gives:

\(4x^2 - 4x + 1 - 9x^2 - 18x - 9 < 0\)

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

Solving the quadratic equation \(-5x^2 - 22x - 8 = 0\) gives the critical values \(x = -4\) and \(x = -\frac{2}{5}\).

The solution to the inequality is \(x < -4\) or \(x > -\frac{2}{5}\).

To solve the inequality \(2|3x - 1| < |x + 1|\), we first consider the critical points where the expressions inside the absolute values are zero: \(3x - 1 = 0\) and \(x + 1 = 0\).

Solving these gives \(x = \frac{1}{3}\) and \(x = -1\). These points divide the number line into intervals.

Consider the intervals: \((-\infty, -1)\), \((-1, \frac{1}{3})\), and \((\frac{1}{3}, \infty)\).

For each interval, remove the absolute values by considering the sign of the expressions:

1. For \(x < -1\), both \(3x - 1\) and \(x + 1\) are negative, so the inequality becomes \(2(-(3x - 1)) < -(x + 1)\).

2. For \(-1 < x < \frac{1}{3}\), \(3x - 1\) is negative and \(x + 1\) is positive, so the inequality becomes \(2(-(3x - 1)) < x + 1\).

3. For \(x > \frac{1}{3}\), both \(3x - 1\) and \(x + 1\) are positive, so the inequality becomes \(2(3x - 1) < x + 1\).

Solving these inequalities gives the critical values \(x = \frac{3}{5}\) and \(x = \frac{1}{7}\).

The solution is \(\frac{1}{7} < x < \frac{3}{5}\).

To solve the inequality \(2 - 5x > 2|x - 3|\), we consider the expression inside the absolute value, \(|x - 3|\), which can be split into two cases:

Case 1: \(x - 3 \geq 0\) (i.e., \(x \geq 3\))

In this case, \(|x - 3| = x - 3\). The inequality becomes:

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

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

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

\(8 > 7x\)

\(x < \frac{8}{7}\)

However, since \(x \geq 3\), there is no solution in this case.

Case 2: \(x - 3 < 0\) (i.e., \(x < 3\))

In this case, \(|x - 3| = -(x - 3) = -x + 3\). The inequality becomes:

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

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

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

\(-4 > 3x\)

\(x < -\frac{4}{3}\)

Thus, the solution is \(x < -\frac{4}{3}\).