Let's break down and solve the two parts of this problem.

Part (a) $6 \leq x^2 + x < 12 + 2x$

We are given a compound inequality:

$6 \leq x^2 + x < 12 + 2x$

Step 1: Break this into two inequalities.

The first inequality is:

$6 \leq x^2 + x$

The second inequality is:

$x^2 + x < 12 + 2x$

Solving the first inequality: $6 \leq x^2 + x$

Rearrange the inequality:

$x^2 + x - 6 \geq 0$

Now, solve $x^2 + x - 6 = 0$ by factoring:

$(x - 2)(x + 3) = 0$

The roots are $x = 2$ and $x = -3$.

The inequality $x^2 + x - 6 \geq 0$ holds in the intervals $(-\infty, -3] \cup [2, \infty)$ because the quadratic opens upwards.

So, the solution to this inequality is:

$x \in (-\infty, -3] \cup [2, \infty)$

Solving the second inequality: $x^2 + x < 12 + 2x$

Rearrange the inequality:

$x^2 + x - 12 - 2x < 0$

Simplify:

$x^2 - x - 12 < 0$

Now, solve $x^2 - x - 12 = 0$ by factoring:

$(x - 4)(x + 3) = 0$

The roots are $x = 4$ and $x = -3$.

The inequality $x^2 - x - 12 < 0$ holds between the roots, so the solution is:

$x \in (-3, 4)$

Combine the results:

From the first inequality, we have $x \in (-\infty, -3] \cup [2, \infty)$, and from the second inequality, we have $x \in (-3, 4)$. The solution is the intersection of these sets:

$x \in [-3, -3] \cup [2, 4) = \{-3\} \cup [2, 4)$

Thus, the solution for part (a) is:

$x \in \{-3\} \cup [2, 4)$

Part (b) $|4x - 1| \geq x - 6$

This is an absolute value inequality. We need to consider two cases based on the definition of absolute value.

Case 1: $4x - 1 \geq x - 6$

Rearrange the inequality:

$4x - x \geq -6 + 1$

$3x \geq -5$

$x \geq -\frac{5}{3}$

Case 2: $-(4x - 1) \geq x - 6$

Simplify:

$-4x + 1 \geq x - 6$

Rearrange the inequality:

$-4x - x \geq -6 - 1$

$-5x \geq -7$

$x \leq \frac{7}{5}$

Combine the results:

From Case 1, we have $x \geq -\frac{5}{3}$, and from Case 2, we have $x \leq \frac{7}{5}$.

Thus, the solution is:

$x \in \left[-\frac{5}{3}, \frac{7}{5}\right]$

Final Answers:

1. For part (a): $x \in \{-3\} \cup [2, 4)$
2. For part (b): $x \in \left[-\frac{5}{3}, \frac{7}{5}\right]$

Do you want further clarification on any of the steps or details?

Here are five related questions to expand your understanding:

1. How do we determine the sign of a quadratic expression to solve inequalities?
2. What is the significance of the intersection when solving compound inequalities?
3. How can you use the graph of a quadratic function to solve inequalities?
4. What changes when solving absolute value inequalities compared to regular inequalities?
5. How do you test the intervals when solving inequalities involving quadratic expressions?

Tip: Always check for extraneous solutions when dealing with absolute value inequalities by substituting the solutions back into the original inequality.