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 :

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

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

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

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