Presumably the idea is to use De L'Hôpital's Rule (see here - particularly Corollary 1).
I could help you further if you have trouble differentiating.
EDIT: Of course, $\cos \pi = -1$, so that the first limit does not exist (denominator diverges). My apologies.
EDIT 2: While I see others are making errors in the application of De L'Hôpital's Rule (HR from now), let me apply it to the second case:
We have that $1-2\cos 0 +\cos 2\cdot 0 = 1 -2 +1=0 = 0^2$, so that HR yields:
$\lim_{x\to 0} \frac{1-2\cos x+\cos2x}{x^2} = \lim_{x\to0} \frac{2\sin x-2\sin2x}{2x}$
We see that it is required to apply HR again, this time yielding:
$\lim_{x\to0} \frac{2\sin x-2\sin2x}{2x} = \lim_{x\to0} \frac{2\cos x-4\cos 2x}{2}$
This latter limit we can simply evaluate by continuity of the involved functions; it follows that the limit is $-1$. In summary, we conclude:
$\lim_{x\to 0} \frac{1-2\cos x+\cos2x}{x^2} = -1$