This comes down to just group theory. Let $Z(G)$ denote the centre of $\rho(G)$.
If $(V, \rho)$ is cyclic, then $\rho(G)/Z(G)$ is cyclic, from which it follows that $\rho(G)$ is abelian. Hence, by Schur's lemma, $(V, \rho)$ is reducible, so a direct sum of one dimensional representations, and Artin's conjecture follows from class field theory.
If $(V, \rho)$ is dihedral, then $\rho(G)/Z(G)$ has an index two subgroup isomorphic to $C_n$. Pulling back this subgroup to $G$, we see that $G$ has an index $2$ subgroup $H$ such that $\rho(H)/Z(G)$ is cyclic. Hence, $\rho(H)$ is abelian.
By Schur's lemma, $\rho|_H$ is therefore a sum of two characters. If $\chi$ is one of these characters, by Frobenius reciprocity
$$(\rho|H, \chi) = (\rho, \mathrm{Ind}_H^G(\chi)).$$
Since $\rho$ is irreducible (e.g. it does not have abelian image), it follows that $\rho\simeq\mathrm{Ind}_H^G(\chi)$.
Conversely, if $\rho \simeq \mathrm{Ind}_H^G(\chi)$ is irreducible, where $H$ is an index two subgroup of $G$, then $H/Z(G)$ is an index two subgroup of $G/Z(G)$. Since $A_4, S_4$ and $A_5$ have no index two subgroups, $G/Z(G)$ must be $D_n$.
You should take a look at Section 4 of this paper, which gives an accessible account of the theory of dihedral Galois representations.
(1): First, let's define some notations. Let's call $G$ the Galois group, $D$ its dihedral image, $C$ its distinguished cyclic subgroup, $H$ its inverse image in $G$. We fix some $s \in G$ whose image in $D$ is the $s$ of the dihedral presentation.
The quotient of $H$ by its scalar matrices (a central subgroup) is $C$ cyclic, so that $H$ is abelian. Then, we can note that $\rho_0$ is given on $H$ by direct sum of two characters $\chi$ and $\chi'$.
We change the basis for $\rho_0$ so that $k(1,0)$, $k(0,1)$ are the lines stable under $H$ (with characters $\chi,\chi'$) and $\rho_0(s)=\begin{bmatrix}0 & 1\\1&0\end{bmatrix}$.
Then we can easily see that $V = kI_2 \oplus k\cdot \mathrm{diag}(1,-1) \oplus A$, where $A$ is the space of antidiagonal matrices. Clearly, $G$ acts trivially on $kI_2$, $H$ acts trivially on the second summand, but $s$ acts by $-1$, so it is the quadratic character $G \rightarrow G/H$.
As for the last summand, it has a canonical basis ($\begin{bmatrix}0 & 1\\0&0\end{bmatrix}$ and its transpose) that $s$ permutes, and $\mathrm{diag}(a,b)$ acts by $\mathrm{diag}(a/b,b/a)$ on this basis, and we recognize the induced representation from $H$ to $G$ by $\chi/\chi'$ (provided that the twist of $\chi/\chi'$ is $\chi'/\chi$, which mostly follows from the dihedral presentation).
(2): I'm interpreting the question as follows: assume $m \neq 2$ and let $n\geq 1$. Why is there an $x \in \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ such that $x$ fixes $\mathbb{Q}(\zeta_{p^n})$ and $\overline{\rho_0(x)} \neq 1$?
Answer: assume that there is no such $x$: then $G_{\mathbb{Q}(\zeta_{p^n})} \subset G_{K_2}$, where $K_2 \subset K_1$ is the subextension such that $Gal(K_1/K_2)$ is mapped by $\rho_0$ to the scalar matrices. In other words, $K_2 \subset \mathbb{Q}(\zeta_{p^n})$. In particular, $K_2/\mathbb{Q}$ is abelian. But the Galois group of $K_2/\mathbb{Q}$ is isomorphic to $D$ (ie is dihedral), so isn't abelian except if $m \leq 2$. We get a contradiction and we're done.
For the edit question, I don't know Wiles' paper very well, but as far as I understand, the goal is to prove that some Galois representations are modular using deformation theory. But usually, one needs some finiteness properties to use deformation theory (see Mazur's 1989 paper in "Galois group over Q" iirc), which are not satisfied by $G_Q=Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, but are satisfied by Galois groups with "restricted ramification" ie $Gal(\mathbb{Q}_{\Sigma}/\mathbb{Q})$.
(exercise/example: use cyclotomic extensions to show that $H^1(G_Q,\mathbb{F}_p):= \mathrm{Hom}(G_Q,\mathbb{F}_p)$ is infinite).
Best Answer
To prove this proposition, we need some facts about Weil groups of number fields. I recommend John Tate's "Number Theoretic Background" for reference.
I will summarise some facts that we need: The Weil group $W_F$ of a number field $F$ is a topological group equipped with a continuous group homomorphism $r: W_F\to G_F$ with a dense image (where $G_F$ denotes the absolute Galois group of $F$). For any finite Galois extension $E/F$, define $W_E:=r^{-1}(G_E)$. Let $W_E^c=\overline{[W_E, W_E]}$ be the closure of the commutator subgroup of $W_E$. Let $W_{E/F}:=W_F/W_E^c$. Then we have an exact sequence $$0\to W_E^{ab}:=W_E/W_E^c\to W_{E/F}\to G_{E/F}\to 0.$$ Also, the natural map $W_F\to \varprojlim W_{E/F}$ is an isomorphism of topological groups.
Let $\rho:W_F\to \text{GL}_2(\mathbb{C})$ be an semisimple continuous representation. If $\rho$ is reducible, then it is of cyclic type. So we may assume that $\rho$ is irreducible. By the no-small-subgroup property of $\text{GL}_2(\mathbb{C})$, $\rho$ factors thorough $W_{E/F}$. Since $W_E^{ab}$ is an abelian normal subgroup of $W_{E/F}$, and $\rho$ is irreducible, the image of $W_E^{ab}$ consists of constant matrices. Hence, the projection $\overline{\rho}:W_{E/F}\to \text{PGL}_2(\mathbb{C})$ factors through $G_{E/F}$, which is finite. Therefore, $\overline{\rho}$ has a finite image. Since $\rho$ is irreducible, the image is an exceptional or a dihedral group. If it is an exceptional group, then $\rho$ itself is of exceptional type. If it is a dihedral group, then a standard representation-theoretic argument can show that $\rho$ itself is of dihedral type.
Note that this is no longer true if $\rho$ is not semisimple. For example, by the definition of the Weil group, there exist a morphism $$f:W_{\mathbb{Q}}^{ab}\simeq \mathbb{Q}^\times\backslash\mathbb{A}_{\mathbb{Q}}^{\times}\simeq \mathbb{R}^{\times}_{>0}\times\prod_{p}\mathbb{Z}_p^\times\twoheadrightarrow\mathbb{R}$$ and $$g\mapsto\begin{pmatrix} 1 & f(g) \\ 0 & 1 \end{pmatrix}$$ is not one of these types.