Let's break down the problem.

Part (a)

We are asked to find the value of $n$ in the equation: $\left( \sqrt[3]{k^{12}} \right)^5 = k^n$

Step 1: Simplify the expression inside the parentheses.

First, focus on simplifying the term inside the cube root: $\sqrt[3]{k^{12}} = k^{12/3} = k^4$ So the equation becomes: $(k^4)^5 = k^n$

Step 2: Apply the power of a power rule.

Using the rule $(a^m)^n = a^{m \cdot n}$, we have: $k^{4 \cdot 5} = k^n$ $k^{20} = k^n$

Step 3: Equate the exponents.

Since the bases are the same, the exponents must be equal: $n = 20$

So, the value of $n$ is $\boxed{20}$.

Part (b)

We are asked to express $\frac{7}{2 - \sqrt{3}}$ in the form $\sqrt{c} + d$, where $c$ and $d$ are integers.

Step 1: Rationalize the denominator.

To eliminate the square root in the denominator, multiply both the numerator and denominator by the conjugate of $2 - \sqrt{3}$, which is $2 + \sqrt{3}$: $\frac{7}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{7(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}$

Step 2: Simplify the denominator.

Use the difference of squares formula to simplify the denominator: $(2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$

So, the expression simplifies to: $\frac{7(2 + \sqrt{3})}{1} = 7(2 + \sqrt{3}) = 14 + 7\sqrt{3}$

Thus, the expression in the form $\sqrt{c} + d$ is: $\boxed{14 + 7\sqrt{3}}$

Would you like any further explanations or clarifications on these steps? Here are some related questions that might help deepen the understanding:

1. What happens if the base of the exponent is negative in part (a)?
2. Can you apply the same rationalization method to other expressions with square roots in the denominator?
3. How do you handle higher powers of square roots in equations like part (b)?
4. What is the significance of the conjugate in rationalizing expressions?
5. How do you solve similar exponentiation problems when the base is not $k$?

Tip: When rationalizing, always multiply by the conjugate to avoid leaving a square root in the denominator!