(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 \(x > 3\).