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 :

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

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

$$l=\frac{1}{3}l+0\Rightarrow l=0$$
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 $\frac16$?