Edit: Here’s a simpler argument for question 1. I’m still assuming the same things (that is, $p$ is unramified for $f$, does not divide $m$, and it does not divide the leading coefficient of $f$). I’m showing a result essentially similar but easier to formulate, namely that whether $f$ splits or not in $\mathbb{F}_q$ depends only on the class of $q$ mod $m$.
Let $K$ be the splitting field of $f$ (as $K/\mathbb{Q}$ is abelian, every subextension is Galois, and thus $K$ is generated by any root of $f$) $L=\mathbb{Q}(e^{2i\pi/m’})$ where $m’$ is the lcm of $m$ and $q-1$.
Let $\mathfrak{p}$ be some prime ideal of $L$ above $p$, $\mathcal{O}_{\mathfrak{p}}$ be the local ring and $k$ be the residual field. Write $f(x)=c(x-x_1)\ldots (x-x_n)$, where $x_i \in \mathcal{O}_{\mathfrak{p}} \cap K$, $c \in \mathbb{Z}_{(p)}^{\times}$.
So $f$ splits in $\mathbb{F}_q$ iff the image of each $x_i$ in $k$ is in $\mathbb{F}_q$, iff for each $1 \leq i \leq n$, the images in $k$ of $x_i$ and $x_i^q$ are the same. Write $q=p^r$, let $\psi \in Gal(L/\mathbb{Q})$ be the Frobenius at $p$, then $f$ splits in $\mathbb{F}_q$ iff for each $1 \leq i \leq n$, the images in $k$ of $x_i$ and $\psi^r(x_i)$ are the same (by definition of $\psi$).
Now, if $1 \leq i \leq n$, $\psi^r(x_i)$ is another root of $f$, hence some $x_j$; $f$ is separable mod $p$ so the images of the $x_i$ in $k$ are pairwise distinct. Therefore, $f$ splits in $\mathbb{F}_q$ iff $\psi^r$ has all the $x_i$ as fixed points, iff $\psi^r$ is in $Gal(L/K)$.
But what is $\psi^r$ in $Gal(L/\mathbb{Q})$ depends only on $\psi^r(e^{2i\pi/m’})=e^{2i\pi p^r/m’}=e^{2i\pi q /m’}$. But $e^{2i\pi q/m’}$ depends only on $q$ mod $m’$. By the Chinese remainder theorem, what $q$ mod $m’$ is depends only on what $q$ mod $m$ is. QED.
Original answer: For question 1, I think the answer is yes as long as $p$ does not divide $m$.
Let $p$ be an unramified prime for $f$, and $q=p^r$. Let $K$ be the splitting field of $f$, so that $K \subset K_0=\mathbb{Q}(e^{2i\pi/m})$.
Then $f$ splits in $\mathbb{F}_q$ iff the residual field of $K$ at a prime over $p$ is contained in $\mathbb{F}_q$. This occurs iff the residual degree at $p$ of $K/\mathbb{Q}$ divides $r$. As $K/\mathbb{Q}$ is unramified at $p$, said residual degree $\delta$ is the cardinality of the decomposition subgroup at $p$ of $Gal(K/\mathbb{Q})$, thus the order of the Frobenius at $p$ in $Gal(K/\mathbb{Q})$.
Thus $\delta$ is the smallest positive integer such that $Frob_p^{\delta} \in Gal(K_0/K)$ (where this Galois group is seen as a subgroup of $Gal(K_0/\mathbb{Q})$ and we are speaking of a Frobenius above $\mathbb{Q}$). But under the identification $Gal(K_0/\mathbb{Q}) \cong (\mathbb{Z}/(m))^{\times}$, $Frob_{\ell}$ is mapped to $\ell$ and $Gal(K_0/K)$ is mapped to the image $H$ of $I$ in $(\mathbb{Z}/(m))^{\times}$, which is thus a subgroup. Therefore, $f$ splits in $\mathbb{F}_q$ iff $r$ is a multiple of the smallest integer $\delta>0$ such that $p^{\delta} \in H$, iff ($H$ is a subgroup) $q \in H$ iff $q \text{ mod } m \in I$.
Let $G = \Gamma$, let $H$ be the decomposition group of $w|v$, so $H$ acts on $K_w$. Then
$$K \otimes_{k} k_v \simeq \mathrm{Ind}^{G}_{H} K_w,$$
and the same is true after passing to units. Hence, by Shapiro's Lemma,
$$H^2(G,(K \otimes_{k} k_v)^{\times}) \simeq H^2(H,K^{\times}_w)$$
But $H$ is identified with $\mathrm{Gal}(K_w/k_v)$, and the result you ask for is then a standard result in the computation of a Brauer group of a local field (Corollary 2, page 131, Serre's notes in Cassels-Frohlich). Note that Serre also shows that this group is generated by $u_{L/K} \in \mathrm{Br}(k_v)$ with invariant $1/d_v \in \mathbf{Q}/\mathbf{Z}$ ("canonically").
Actually, for a more direct reference, see Corollary 7.4(b) of Tate's notes in the next chapter of the same book on page 177.
As to the last question, the map is as you expect; viewing the first element as $1/d_v$ inside $\mathbf{Q}/\mathbf{Z}$ and the latter as $1/d$ inside $\mathbf{Q}/\mathbf{Z}$, it is induced by the identity map from $\mathbf{Q}/\mathbf{Z}$ to itself, so in your setting is the injective map sending $1/d_v$ to $(d/d_v) \cdot 1/d$. This is because $H^2(K/k,C_K)$ is more or less the $d$ torsion in $\mathbf{Q}/\mathbf{Z}$ which is the "final term" in the standard short exact sequence of class field theory:
$$0 \rightarrow \mathrm{Br}(k_v) \rightarrow \bigoplus_{v} \mathrm{Br}(k_v) \rightarrow \mathbf{Q}/\mathbf{Z} \rightarrow 0.$$
For a precise reference, combine the discussion in Consequence 9.6 of page 185 (of Tate's notes) with diagram (9) in section 11, bottom of page 195.
To make sure you look at the right edition of Cassels-Frohlich I am using the one with this page numbering:
https://www.math.arizona.edu/~cais/scans/Cassels-Frohlich-Algebraic_Number_Theory.pdf
Best Answer
There is a standard exact sequence $1 \to \pi_1(X,a) \to \pi_1(X_0,a) \to Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q) \to 1,$ given by covariant functoriality of $\pi_1$.
How can one think about $\pi_1(X_0,a)$? Well, it is a quotient of the absolute Galois group of the function field $K(X_0)$ of $X_0$, and can be thought of as the Galois group of the maximal Galois extension of $K(X_0)$ which is everywhere unramified. Thus $\pi_1(X_0,a)^{ab}$ is the Galois group of the maximal abelian extension of $K(X_0)$ which is everywhere unramified.
Class field theory gives a description of this everywhere unramified abelian extension in terms of class groups/ideles. When you sort it out in this context, you will find that $\pi_1(X_0,a)^{ab}$ gets identified with the profinite completion of the $\mathbb{F}_q$-rational points of $Pic(X_0)$. Concretely, $Pic^0(X_0)(\mathbb F_q)$ is finite, and the degree map induces an exact sequence $$0 \to Pic^0(X_0)(\mathbb F_q) \to Pic(X_0)(\mathbb F_q) \to \mathbb Z \to 0.$$ Taking profinite completions just replaces $\mathbb Z$ by $\hat{\mathbb Z}$.
Now going back to the exact sequence at the beginning of the answer, since $Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q)$ is abelian, there is an induced sequence $$\pi_1(X,a) \to \pi_1(X_0,a)^{ab} \to Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q) \to 1.$$ Since $Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q)$ is a copy of $\hat{\mathbb{Z}}$, generated by $Frob_q$, we may rewrite this as $$\pi_1(X,a) \to \pi_1(X_0,a)^{ab} \to \hat{\mathbb Z} \to 0.$$
Now one checks that under the identification of $\pi_1(X_0,a)^{ab}$ and the profinite completion of $Pic(X_0)(\mathbb F_q)$, the map to $\hat{\mathbb Z}$ of this last exact sequence is the same as the degree map.
Comparing exact sequences, we see that the image of $\pi_1(X,a)$ in $\pi_1(X_0,a)^{ab}$ is equal to $Pic^0(X_0)(\mathbb{F}_q)$, which is what is being asserted.
[Note: The reason that Weil groups are invoked are that, in this context, passing from Galois groups to Weil groups simply replaces $\hat{\mathbb Z}$ by $\mathbb Z$. So $\pi_1(X_0,a)^{ab}$ is a Galois group, while $Pic(X_0)(\mathbb{F}_q)$ is the corresponding Weil group. The latter has a surjection to $\mathbb Z$, while the former has a surjection to $\hat{\mathbb Z}$. If you compute the quotient of the idele class group pertaining to the maximal abelian everywhere unramified extension, you will naturally get $Pic(X_0)(\mathbb F_q)$, i.e. the Weil group. In the function field case the Artin reciprocity map from quotients of the idele class group to abelian Galois groups is injective, with image being the Weil group.
By constrast, in the number field case the reciprocity map is surjective, but has a kernel, coming from the contribution at the archimedean primes.
A careful discussion of Weil groups, including the distinction between the number field and function field cases, can be found in Tate's Numer theoretic background article in the Corvalis proceedings.]