Question: I understand why $\cos(\theta) = \cos(-\theta)$ but I don't know what the specific property this question is asking for is - PreCalc homework.

Likewise, another question is:
$\sin(-\theta) = -\sin(\theta)$ which means that the sine function is (blank)
which I again understand why it is but I don't know the property.

Answer: 
you can prove that the limit exist bu several way

so to find the limit without l'hopital :

$$l=\lim\_{x\to 0}\frac{x-\sin x}{x^2}\stackrel{x\to3x}{=}\frac{1}{9}\lim\_{x\to 0}\frac{3x-3\sin x+4\sin^3x}{x^2}$$

$$l=\frac{1}{3}\lim\_{x\to 0}\frac{x-\sin x}{x^2}+\frac{4}{9}\lim\_{x\to 0}\frac{\sin^3x}{x^2}$$

$$l=\frac{1}{3}l+0\Rightarrow l=0$$
The "right" thing to use is the Taylor series, but we will work our way around that. We also avoid L'Hospital's Rule. By the Mean Value Theorem, for any $x\ne 0$ there is a $c\_x$ between $0$ and $x$ such that 
$$\frac{x-\sin x}{x}=1-\cos c\_x=2\sin^2(c\_x/2).$$
Thus
$$\frac{\sin x-x}{x^2}=\frac{2\sin^2(c\_x/2)}{x}.$$
Suppose $|x|$ is small non-zero. Since $|c\_x/2|\lt |x|/2$ we have
$$0\lt \frac{2\sin^2(c\_x/2)}{|x|}\lt \frac{x^2/2}{|x|}\lt |x|/2.$$
The right-hand side approaches $0$ as $x\to 0$, so our limit is $0$.