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I'm having the hardest time trying to find these and need some help.

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For any abelian group $A$, you may take the free abelian group $F = \mathbb{Z}\left$ generated by the elements of $A$ and the corresponding epimorphism $F\twoheadrightarrow A$. (Instead of all elements of $A$, you may also consider its generators, if that makes things easier.) Then its kernel $K$ is also free, being a subgroup of a free abelian group (that's a very special property of $\mathbb{Z}$ as a base ring). So $$0 \to K \to F \to A \to 0$$ is a projective (actually, free) resolution.

For instance, for $\mathbb{Z} \left[\frac{1}{p}\right]$, take the free abelian group generated by formal symbols $\frac{1}{p^i}$ for $i = 0,1,2,3,\ldots$ This is the group of formal finite sums $\sum_{i\ge 0} n_i\,\frac{1}{p^i}$ with coefficients $n_i \in \mathbb{Z}$ (where $n_i \ne 0$ only for finitely many $i$). This group surjects to $\mathbb{Z} \left[\frac{1}{p}\right]$ (by sending a formal sum to the corresponding element of $\mathbb{Z} \left[\frac{1}{p}\right]$). This surjection has some very nontrivial kernel, but it's a free abelian group.

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    Thanks so much. I'm a little unfamiliar with the notation you're using. Can you please write down explicitly what $F$ would be in the above examples in particular? – 2017-01-19
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    The free abelian group generated by the elements of $A$. I think another common notation is $\mathbb{Z}^{(A)}$. It is isomorphic to the direct sum $\bigoplus_{a\in A} \mathbb{Z}$. – 2017-01-19