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: 
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 $0$.
You can use l'Hopital's rule if you know how to take derivatives. After checking that the fraction is naively $0/0$, differentiate the top of the fraction, and the bottom of the fraction.
$$
\frac{1-\cos x}{2x}$$
Still $0/0$? Then do it again:
$$
\frac{\sin x}{2} \to 0$$

SO the answer will be zero. Are you sure the problem did not read $$\lim\_{x \to 0}\frac{x- \sin{x}}{x^3}$$, which requires three applicatoins and gives an answer of $\frac16$?