Find the value of k for which the limit of the piecewise function exists and evaluate the given limit using L'Hôpital's Rule.

$$
\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (kx^2 + 1)
$$
Substitute $x=2$:
$$
k(2)^2 + 1 = 4k + 1
$$

$$
\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - k)
$$
Substitute $x=2$:
$$
3(2) - k = 6 - k
$$

$$
4k + 1 = 6 - k
$$
$$
4k + k = 6 - 1
$$
$$
5k = 5
$$
$$
k = 1
$$
The value of $k$ is $\boxed{1}$.

$$
\frac{e^0 - 0 - 1}{0^2} = \frac{1 - 0 - 1}{0} = \frac{0}{0}
$$
This is an indeterminate form, so we apply L'Hôpital's Rule.

$$
\lim_{x \to 0} \frac{\frac{d}{dx}(e^x - x - 1)}{\frac{d}{dx}(x^2)} = \lim_{x \to 0} \frac{e^x - 1}{2x}
$$

$$
\frac{e^0 - 1}{2(0)} = \frac{1 - 1}{0} = \frac{0}{0}
$$
This is still an indeterminate form, so we apply L'Hôpital's Rule again.

$$
\lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(2x)} = \lim_{x \to 0} \frac{e^x}{2}
$$

$$
\frac{e^0}{2} = \frac{1}{2}
$$
The limit is $\boxed{\frac{1}{2}}$.

The limit is of the form $\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^{bx} = e^{ab}$.

In this case, $a=3$ and $b=2$.

$$
e^{(3)(2)} = e^6
$$
The limit is $\boxed{e^6}$.

For any $x \ne 0$, the value of $\cos\left(\frac{1}{x}\right)$ is always between $-1$ and 1.
$$
-1 \le \cos\left(\frac{1}{x}\right) \le 1
$$

Multiply the inequality by $x^2$. Since $x^2 \ge 0$, the direction of the inequalities remains the same.
$$
-x^2 \le x^2 \cos\left(\frac{1}{x}\right) \le x^2
$$

$$
\lim_{x \to 0} (-x^2) = -(0)^2 = 0
$$
$$
\lim_{x \to 0} (x^2) = (0)^2 = 0
$$

Since both bounding functions approach 0 as $x \to 0$, the limit of the middle function must also be 0.  
The limit is $\boxed{0}$.

$$
|x+2| = \begin{cases} x+2 & \text{if } x+2 \ge 0 \implies x \ge -2 \\ -(x+2) & \text{if } x+2 < 0 \implies x < -2 \end{cases}
$$

As $x \to -2^-$ (meaning $x$ is slightly less than $-2$), $x+2 < 0$, so $|x+2| = -(x+2)$.
$$
\lim_{x \to -2^-} \frac{-(x+2)}{x+2} = \lim_{x \to -2^-} (-1) = -1
$$