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

Expanding both sides, we have:

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

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

Setting up the inequality:

\(x^2 + 4ax + 4a^2 > 9x^2 - 18ax + 9a^2\)

Simplifying, we get:

\(-8x^2 + 22ax - 5a^2 > 0\)

Solving the quadratic equation \(-8x^2 + 22ax - 5a^2 = 0\) using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a = -8\), \(b = 22a\), \(c = -5a^2\).

\(x = \frac{-22a \pm \sqrt{(22a)^2 - 4(-8)(-5a^2)}}{2(-8)}\)

\(x = \frac{-22a \pm \sqrt{484a^2 - 160a^2}}{-16}\)

\(x = \frac{-22a \pm \sqrt{324a^2}}{-16}\)

\(x = \frac{-22a \pm 18a}{-16}\)

\(x = \frac{-22a + 18a}{-16}\) or \(x = \frac{-22a - 18a}{-16}\)

\(x = \frac{-4a}{-16} = \frac{1}{4}a\) or \(x = \frac{-40a}{-16} = \frac{5}{2}a\)

Thus, the solution to the inequality is \(\frac{1}{4}a < x < \frac{5}{2}a\).

To solve the inequality \(|4x + 3| > |x|\), we consider two cases based on the definition of absolute value.

Case 1: \(4x + 3 > x\)

Simplify to get \(3x + 3 > 0\), which gives \(x > -1\).

Case 2: \(4x + 3 < -x\)

Simplify to get \(5x < -3\), which gives \(x < -\frac{3}{5}\).

Combining both cases, the solution is \(x < -1\) or \(x > -\frac{3}{5}\).

(a) The graph of \(y = |2x + 3|\) is a V-shaped graph with the vertex at the point where \(2x + 3 = 0\), which is \(x = -\frac{3}{2}\). The graph is symmetric about the vertical line through the vertex.

(b) To solve \(3x + 8 > |2x + 3|\), consider the two cases for the absolute value:

Case 1: \(3x + 8 > 2x + 3\)

Simplifying gives \(x > -5\).

Case 2: \(3x + 8 > -(2x + 3)\)

Simplifying gives \(3x + 8 > -2x - 3\), which leads to \(5x > -11\) and \(x > -\frac{11}{5}\).

The solution is the intersection of these two cases, which is \(x > -\frac{11}{5}\).

To solve \(|x - 2| = \left|\frac{1}{3}x\right|\), consider the cases for the absolute values.

Case 1: \(x - 2 = \frac{1}{3}x\)

Rearrange to get: \(x - \frac{1}{3}x = 2\)

\(\frac{2}{3}x = 2\)

\(x = 3\)

Case 2: \(x - 2 = -\frac{1}{3}x\)

Rearrange to get: \(x + \frac{1}{3}x = 2\)

\(\frac{4}{3}x = 2\)

\(x = \frac{3}{2}\)

Thus, the solutions are \(x = 3\) or \(x = \frac{3}{2}\).

To solve \(|4x - 1| = |x - 3|\), consider the cases for the absolute values.

Case 1: \(4x - 1 = x - 3\)

Solve for \(x\):

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

\(3x = -2\)

\(x = -\frac{2}{3}\)

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

Solve for \(x\):

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

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

\(5x = 4\)

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

Thus, the solutions are \(x = -\frac{2}{3}\) and \(x = \frac{4}{5}\).