I'll get you started for (1):
For each prime $p$, we have three possibilities: $p$ divides $d$ and hence ramifies in $\mathcal{O}_K$, or $p \nmid d$ in which case $p$ splits completely or remains inert. In either case, picking a prime $Q$ in $\mathcal{O}_K$ above $p\mathbb{Z}$ supplies us with a residual field $F_Q=\mathcal{O}_K/Q$ over $\mathbb{F}_p$. The 'local' Galois group $\text{Gal}(F_Q/\mathbb{F}_p)$ embeds into the 'global' Galois group $\text{Gal}(K/\mathbb{Q})$ at the unramified primes. Since $|\text{Gal}(F_Q/\mathbb{F}_p)| = [F_Q : \mathbb{F}_p]=\log_p|F_Q|$, this embedding $\text{Gal}(F_Q/\mathbb{F}_p) \hookrightarrow \text{Gal}(K/\mathbb{Q})$ is the trivial map if and only if $F_Q/\mathbb{F}_p$ is a trivial extension, which is the case if and only if $p$ splits. In any case, the image of the canonical generator of $\text{Gal}(F_Q/\mathbb{F}_p)$ (which is cyclic) is defined to be the Frobenius at $p$, $\text{Frob } p$. It depends only up to conjugation on the choice of $Q$ lying over $p$.
In the case of a quadratic extension, $\text{Gal}(K/\mathbb{Q})=\{1, \sigma\}$, and we get the following characterization of the Frobenius:
$\text{Frob }p= \begin{cases}\sigma & \text{if }p \text{ is inert} \\ 1 & \text{if }p\text{ splits}\end{cases}$
Now if we are given a representation $\rho: \text{Gal}(K/\mathbb{Q}) \to \text{GL}(V)$, we get a map on the primes of $\mathbb{Z}$ to $\text{GL}(V)$ by composing with the Frobenius: $p \mapsto \rho(\text{Frob p})$. This map is well-defined up to conjugation in $\text{GL}(V)$. It extends by multiplicativity to all positive integers - this is the Artin map.
In the case where the global Galois group is abelian, or the representation one-dimensional, everything is well-defined. This is the case you are dealing with.
There is only one nontrivial irreducible representation $\rho$ of the cyclic group on two elements, and it's the obvious map to $\{\pm 1\}$. This is the origin of the Dirichlet character associated to a quadratic extension: it is the Artin map associated to the representation $\rho$.
Now how do we bridge the gap with the usual definition of a Dirichlet character, as an element of the dual of $\mathbb{Z}/d\mathbb{Z}$? This is the content of the theorem of Quadratic Reciprocity.
For (2), recall that the Dedekind zeta-function of a number field always has a simple pole at $s=1$. Also, $\zeta(s)$ has a pole at $s=1$. Comparing orders at $s=1$ on both sides yields the result (which has immediate implications!)