$\newcommand{\mf}[1]{\mathfrak{#1}}$
$\newcommand{\oO}{\mathcal{O}}$
$\newcommand{\ZZ}{\mathbb{Z}}$
$\newcommand{\Gal}{\text{Gal}}$
$\newcommand{\CC}{\mathbb{C}}$
$\newcommand{\QQ}{\mathbb{Q}}$
For (1), the definition goes like this: Suppose you have a galois extension of number fields $L/K$, then you have an extension of rings of integers $\oO_L/\oO_K$. The galois group $\Gal(L/K)$ acts as automorphisms on $L$ that fix $K$. Hence, it acts on $\oO_L$ and fixes $\oO_K$. Thus, for a prime $\mf{p}$ in $\oO_K$, and a prime $\mf{q}$ in $\oO_L$ containing $\mf{p}$, any $\sigma\in\Gal(L/K)$ must fix $\mf{p}$, and hence it must send $\mf{q}$ to another prime $\mf{q}'$ in $\oO_L$ which also contains $\mf{p}$ - ie, lies above $\mf{p}$. Thus, you may consider the subgroup of $\Gal(L/K)$ which fixes $\mf{q}$ - this is a subgroup of $\Gal(L/K)$, and is called the decomposition group $D_{\mf{q}/\mf{p}}$. In fact, $\Gal(L/K)$ acts transitively on the primes of $\oO_L$ lying above $\mf{p}$, and thus if $\sigma\in\Gal(L/K)$ sends $\mf{q}\mapsto\mf{q}'$, then $D_{\mf{q'}/\mf{p}} = \sigma(D_{\mf{q}/\mf{p}})\sigma^{-1}$, so the decomposition groups for primes lying above $\mf{p}$ are all conjugate (but still in general they are different subgroups).
Now, since $D_{\mf{q}/\mf{p}}$ fixes $\mf{q}$, it determines a well-defined action on $\oO_L/\mf{q}$, which is naturally seen to be a finite field extension of $\oO_K/\mf{p}$. In general, if you have a surjection of groups or rings $A\twoheadrightarrow B$, then an automorphism of $A$ descends to an automorphism of $B$ if and only if it fixes the kernel.
The elements of $D_{\mf{q}/\mf{p}}$ which induce the identity on $(\oO_L/\mf{q})/(\oO_K/\mf{p})$ are precisely the automorphisms $\sigma\in D_{\mf{q}/\mf{p}}$ such that $\sigma(x)\equiv x\mod \mf{q}$ for all $x\in\oO_L$. These automorphisms form a subgroup called the inertia group $I_{\mf{q}/\mf{p}}$.
Thus, there are inclusions $\{1\}\subset I_{\mf{q}/\mf{p}}\subset D_{\mf{q}/\mf{p}}\subset\Gal(L/K)$, which by galois theory corresponds to extensions of fields $L$ over $L^{I_{\mf{q}/\mf{p}}}$ over $ L^{D_{\mf{q}/\mf{p}}}$ over $K$. Going right to left, the prime $\mf{p}$ of $K$ decomposes completely in $L^{D_{\mf{q}/\mf{p}}}$, remains inert until $L^{I_{\mf{q}/\mf{p}}}$, and ramifies from there until $L$.
For (3), imagine the horizontal parabola given by, say, $x = y^2$, projecting onto the $x$-axis. At $x = 1$, you can solve the equation to find that $(1,1)$ and $(1,-1)$ are both points on the parabola which map to 1. Indeed, in general for $x = a$, the preimage of $a$ is $\{(a,\sqrt{a}),(a,-\sqrt{a})\}$. Thus, the preimage almost always has size 2, with one exception, namely when $a = 0$ (if $a < 0$ then the picture becomes deficient, but if you work over $\mathbb{C}$, you still have two (imaginary) solutions). Thus, this is an example of a finite map (ie, map with finite fibers, in this case generically of size 2, so you call the map a degree 2 map), which is ramified above $x = 0$, in the sense that above $x = 0$, there are fewer preimages than normal (in this case 1). In general this is what "ramification" means in geometry. An unramified map is one where the preimage (ie, fiber) above any point are all finite and have the same size.
What does this have to do with primes? Well, one of the great insights of Grothendieck is that prime ideals are kind of like points! For example, if you consider the ring $\CC[x]$, what are its prime ideals? Well, there is the zero ideal $(0)$, and then by the fundamental theorem of algebra, the other primes are all of the form $(x - a)$, where $a\in\CC$. Thus, we get a bijection between the maximal prime ideals of $\CC[x]$ and the points of $\CC$ by sending $(x-a)\mapsto a$. The $(0)$ ideal is kind of special - it's called the generic point, and in a sense represents the entirety of $\CC$.
Note that the rings $\CC[x]$, $\oO_K$ are very similar! They're both Dedekind domains, and in fact $\CC[x]$ is even a PID (class number 1). The fraction field of $\CC[x]$ is $\CC(x)$ and is analogous to $K$. A galois extension of $\CC(x)$ looks like $\CC(x)[y]/f(y)$ and is analogous to the extension $L/K$. Then, $\oO_L$ is analogous to the integral closure of $\CC[x]$ in $\CC(x)[y]/f(y)$.
If you take our example with the parabola, you could set $f(y) = y^2 - x$, then $L' := \CC(x)[y]/(y^2-x)$ is a quadratic extension of $K' := \CC(x)$. It is galois, with galois group of order 2 (hence isomorphic to $\mathbb{Z}/2\mathbb{Z}$, and is generated by the automorphism which fixes $\CC(x)$ and sends $y\mapsto -y$. The integral closure of $\oO_{K'} := \CC[x]$ inside $L'$ in this case is just $\oO_{L'} := \CC[x,y]/(y^2-x)$.
For the prime $\mf{p}_a := (x-a)$ of $\CC[x]$, then there are two primes of $\oO_{L'}$ lying above $\mf{p}_a$, namely $(y-\sqrt{a}),(y+\sqrt{a})$, which are different primes, unless $a = 0$, in which case they are the same.
Exercise: Compute the inertia and decomposition groups for primes of $\oO_{L'}$ lying above $(x - 1),(x-2)$ and $(x-0) = (x)$. You'll find that in the first two cases the decomposition group and inertia group are both trivial. In the second case, the decomposition group and inertia group are both the whole galois group. Note that in this situation primes never remain inert, simply because all the residue fields of $\CC[x]$ are just $\CC$, which is algebraically closed. Note that in each case the galois group acts transitively on the primes of $\oO_{L'}$ lying above a prime of $\oO_{K'}$.
Exercise: Play the same game above with $\CC[x]$ replaced by $\QQ[x]$. Note that $\QQ[x]$ is still a Dedekind domain (in fact a PID), so all the definitions of decomposition/inertia still make sense. You should find that $(x-1)$ splits, $(x-2)$ remains inert, and $(x-0) = (x)$ is ramified.
Remark: In the language of schemes, the topological space $\CC$, viewed as a scheme, is Spec $\CC[x]$, where Spec $R$ for a ring $R$ is a topological space whose underlying set is the set of prime ideals of $R$. The topology however is called the Zariski topology and in the case of Spec $\CC[x]$, does not agree with the complex topology on $\CC$. However, the cool thing here is that the traditionally geometric notions of dimension, smoothness, connectedness, compactness, fundamental groups, picard groups, tangent spaces, differential forms,...etc can all be defined purely algebraically for RINGS. From this viewpoint, the ring of integers of an algebraic number field is a smooth connected non-compact curve. $\ZZ$ has trivial fundamental group, since all rings of integers are ramified over $\ZZ$ (this follows I believe from the fact that the discriminant of a number field is never a unit, and comes from the geometry of numbers (NOT algebraic geometry)).
Best Answer
Note that if $k'/k$ is any finite separable extension of fields, then $k' \otimes_k \bar{k} \cong \bar{k}^{[k' : k]}$. So, given an extension of number fields $L/K$, a nonzero prime ideal $\mathfrak{p} \subseteq \mathcal{O}_K$ with residue field $k = \mathcal{O}_K/\mathfrak{p}$ is unramified in the extension $\mathcal{O}_L/\mathcal{O}_K$ if and only if the geometric fiber $\operatorname{Spec}(\mathcal{O}_L/\mathfrak{p} \otimes_{k} \bar{k})$ (that is, the base change of the fiber, which is a $k$-scheme, to the algebraic closure $\bar{k}$) has exactly $[L : K]$ closed points.
This exactly mirrors the situation for branched coverings of Riemann surfaces, where a branch point is exactly one where the fiber has fewer points than the degree of the covering map. The difference is that since we're dealing with non-algebraically-closed fields, we might have to extend the residue field in order to "see" the full size of the fiber.
From this perspective, here are geometric interpretations of inertia degree and ramification index of a prime $\mathfrak{b} \in \operatorname{Spec} \mathcal{O}_L$: