Let $R = \mathcal{O}_{K}$. As you note, $R = \mathbb{Z}[\sqrt{2}] \cong \mathbb{Z}[X]/\langle X^{2}-2 \rangle$. Note that for any prime $p$, $R/pR \cong \mathbb{Z}[X]/\langle p, X^{2}-2 \rangle \cong (\mathbb{Z}/p\mathbb{Z})[X]/\langle X^{2}-2 \rangle$. (In the last isomorphism, I am being a bit sloppy and identifying $X^{2}-2$ with its residue class in $(\mathbb{Z}/p\mathbb{Z})[X]$.)
The factorization of $p$ into a product of prime ideals of $R$ can consequently be understood in terms of how the polynomial $X^{2}-2$ factors over $\mathbb{Z}/p\mathbb{Z}$. There is a general theorem at work here, namely Dedekind's theorem, but you can attack things quite directly. (I would personally view this problem as a nice way to see how Dedekind's theorem works in a simple case.) Here are some hints for each part:
(A) See if you can prove that $\sqrt{2}R$ is a prime ideal; it shouldn't be hard from there to show that $2R = (\sqrt{2}R)^{2}$. (Hint: $R/\sqrt{2}R \cong \mathbb{Z}[X]/\langle X, X^{2}-2\rangle \cong \ldots$). This reflects the fact that $R/2R \cong (\mathbb{Z}/2\mathbb{Z})[X]/\langle X^{2} \rangle$.
(B), (C): The conditions on $p$ here control whether $2$ is a quadratic residue mod $p$, i.e. whether $X^{2}-2$ splits over $\mathbb{Z}/p\mathbb{Z}$. You should try to show that in (B), $2$ is NOT a quadratic residue mod $p$, and so $X^{2}-2$ is irreducible over $\mathbb{Z}/p\mathbb{Z}$, in which case $pR$ is prime. Likewise, in (C), $2$ IS a quadratic residue mod $p$. Since $p$ is odd, $X^{2}-2$ splits into distinct factors over $\mathbb{Z}/p\mathbb{Z}$. Try to use this factorization (and maybe the Chinese Remainder Theorem) to construct the prime factorization of $pR$.
