Question 4.c asks to find the values of k for which the function g(x) is continuous at x=2.

For $x \le 2$, $g(x) = k^2 x^2$.
Substituting $x=2$:

$$g(2) = k^2 (2)^2 = 4k^2$$

The function is defined at $x=2$.

For $x < 2$, $g(x) = k^2 x^2$.

$$\lim_{x \to 2^-} g(x) = \lim_{x \to 2^-} k^2 x^2 = k^2 (2)^2 = 4k^2$$

For $x > 2$, $g(x) = (1-k)x$.

$$\lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} (1-k)x = (1-k)(2) = 2 - 2k$$

For $g(x)$ to be continuous at $x=2$, we must have:

$$\lim_{x \to 2^-} g(x) = \lim_{x \to 2^+} g(x) = g(2)$$

So, we set the expressions from Step 1, 2, and 3 equal to each other:

$$4k^2 = 2 - 2k$$

Rearrange the equation into a standard quadratic form $ax^2 + bx + c = 0$:

$$4k^2 + 2k - 2 = 0$$

Divide the entire equation by 2 to simplify:

$$2k^2 + k - 1 = 0$$

Factor the quadratic equation:

$$2k^2 + 2k - k - 1 = 0$$ $$2k(k+1) - 1(k+1) = 0$$ $$(2k-1)(k+1) = 0$$

This gives two possible values for $k$:

$$2k-1 = 0 \implies 2k = 1 \implies k = \frac{1}{2}$$ $$k+1 = 0 \implies k = -1$$

The values of $k$ for which $g(x)$ is continuous at $x=2$ are $\boxed{k = \frac{1}{2} \text{ and } k = -1}$.