In Cassels and Frohlich (Algebraic Number Theory) Exercise 1, one is asked to derive some properties of the power residue symbol. It begins by stating the following:
Let $m$ be a fixed natural number and $k$ a fixed global field containing group $\mu_m$ of $m$th roots of unity. Let $S$ denote the set of primes of $k$ consisting of the archimedean ones and those dividing $m$.
If a prime ideal $\frak p$ in $\mathcal{O}_k$ does note divide $m$ is that equivalent to saying it is prime to $m$? (where I mean that $\frak p$ is prime to $m$ if $\frak p$ does not divide $m\mathcal{O}_k$)
Given a number field $K$, its ring of integers $\mathcal{O}_K$ is a Dedekind domain, which means that every ideal has a unique factorization into prime ideals. Given $m \in \mathbb{Z}$, then $m \mathcal{O}_K$ is an ideal of $\mathcal{O}_K$, hence has a prime factorization: $$ m \mathcal{O}_K = \mathfrak{p_1}^{e_1} \cdots \mathfrak{p_t}^{e_t} $$ for some prime ideals $\mathfrak{p_1}, \ldots, \mathfrak{p_t}$ of $\mathcal{O}_K$ and some positive integers $e_1, \ldots, e_t$. These prime ideals are said to divide $m$.
Another way to think about divisibility is by containment. Given ideals $\mathfrak{a}$ and $\mathfrak{b}$ of a Dedekind domain, $\mathfrak{a}$ divides $\mathfrak{b}$ iff $\mathfrak{a} \supseteq \mathfrak{b}$, i.e., to contain is to divide. Note that $$ m \mathcal{O}_K = \mathfrak{p_1}^{e_1} \cdots \mathfrak{p_t}^{e_t} \subseteq \mathfrak{p}_i $$ for $i = 1, \ldots, t$, so this agrees with the previous notion of divisibility.
This allows us to answer your second question. Suppose $\mathfrak{p}$ does not divide $m \mathcal{O}_K$, so $\mathfrak{p} \not \supseteq m \mathcal{O}_K$. Then $\mathfrak{p} + m \mathcal{O}_K \supsetneq \mathfrak{p}$ is strictly larger than $\mathfrak{p}$. But prime ideals are maximal in a Dedekind domain, so this means that $\mathfrak{p} + m \mathcal{O}_K = \mathcal{O}_K$. Then $\mathfrak{p}$ and $m \mathcal{O}_K$ are comaximal, which is a generalization of being relatively prime. (See Bézout's Lemma.)