Let's analyze and solve the questions one by one.

Question 1

The problem gives us a function: $f(x) = 3x^2 - 2$ and asks for the limit representing the derivative at $x = 2$: $\lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h} \quad \text{at } x = 2.$

To solve this:

1. Calculate $f'(x)$, the derivative of $f(x)$. $f(x) = 3x^2 - 2$ The derivative, $f'(x)$, is: $f'(x) = \frac{{d}}{{dx}}(3x^2 - 2) = 6x.$

2. Substitute $x = 2$ to find the slope at this point: $f'(2) = 6 \cdot 2 = 12.$

Thus, the outcome of the limit (and the slope of the tangent line at $x = 2$) is 12, so the answer is (b) 12.

Question 2

The problem asks for the slope of the function at $x = 5$ for: $f(x) = 6\sqrt{2x - 2}.$

To find this slope:

1. Rewrite $f(x)$ to make it easier to differentiate: $f(x) = 6(2x - 2)^{1/2}.$

2. Use the chain rule to differentiate: $f'(x) = 6 \cdot \frac{1}{2} (2x - 2)^{-1/2} \cdot 2 = \frac{6}{\sqrt{2x - 2}}.$

3. Substitute $x = 5$: $f'(5) = \frac{6}{\sqrt{2 \cdot 5 - 2}} = \frac{6}{\sqrt{10 - 2}} = \frac{6}{\sqrt{8}} = \frac{6}{2\sqrt{2}} = \frac{3}{\sqrt{2}}.$

Since none of the options match $\frac{3}{\sqrt{2}}$, the answer is (e) none of these.

Would you like further clarification on any part of the solution?

Here are some related questions to expand your understanding:

1. How would you determine the derivative of a general quadratic function?
2. What are some alternative methods to compute the slope at a point?
3. How can limits help in defining the tangent line for any given function?
4. Why is it essential to rewrite functions for easier differentiation?
5. How does the chain rule apply in problems involving square roots?

Tip: For functions involving square roots, rewriting them in fractional exponents can simplify differentiation.