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: 
L'Hopital is the easy way to do this, but here's another approach, using only $(\sin x)'=\cos x$ and $\sin x\lt x$ (for $x$ positive):

Since ${x-\sin x\over x^2}$ is an odd function, it suffices to consider the one-sided limit with $x\gt0$. We have

$$0\le{x-\sin x\over x^2}={1\over x^2}\int\_0^x(1-\cos u)du={1\over x^2}\int\_0^x{\sin^2u\over1+\cos u}du\le{1\over x^2}\int\_0^x\sin^2udu\\={1\over x^2}\int\_0^xu^2{\sin^2u\over u^2}du\le{1\over x^2}\int\_0^xu^2du={1\over x^2}\cdot{1\over3}x^3={1\over3}x\to0\text{ as }x\to0^+$$
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 $\frac16$?