Let $K/\mathbb{Q}_p$ be a finite extension with ring of integers $R$ and residue field $k$. Let $A/K$ be an abelian variety with Neron model $\mathcal{A}/R$. We denote by $\tilde{\mathcal{A}}/k$ the special fiber and by $\mathcal{A}^0$ - the connected component of $\mathcal{A}$.

1. Is it true that $\mathcal{A}^0(R)[n] \hookrightarrow \tilde{\mathcal{A}^0}(k)$ for $p \nmid n$? What is the right reference for this fact?

This is discussed here in case that $A$ has good reduction. Liu's answer in that topic addresses the general case:

(...) when A is not an abelian scheme (for example, when it is the Néron model of an abelian variety over $\mathbb{Q}_p$ with not necessary good reduction), then for any n prime to p, the kernel $A[n]$ is still étale over $\mathbb{Z}_p$ (because the tangent map at 0 of the multiplication by n is just multiplication by n for any commutative algebraic group), but not necessarily finite. There is a biggest closed subscheme H of $A[n]$ which is étale and finite over $\mathbb{Z}_p$. The reduction map on H is injective (see Pete's proof). The generic fiber of H corresponds to the points of the generic fiber of $A[n]$ having specialization mod p. (...)

I don't really understand the last sentence - what does it mean for a point to have a specialization mod $p$? Isn't it true that every point in $A(K)$ corresponds to a point in $\mathcal{A}(R)$ and every point in $\mathcal{A}(R)$ can be reduced to $\mathcal{A}(k)$?

2. Is it possible to prove the "$(1) \Rightarrow (2)$" implication of Neron-Ogg-Shafarevich criterion for abelian varieties in the language of Silverman's Arithmetic of Elliptic Curves? This would be my proposition:

Theorem If $A/K$ has good reduction, then inertia $I_K$ acts trivially on $A[n]$ for $p \nmid n$.

Proof: Let $L = K(A[n])$ and let $R', k'$ be the ring of integers and residue field of $L$. Let $\mathcal{A}/R'$ be the Neron model of $A_L$. Then any point $P \in A[n]$ corresponds to a point $\mathcal{P} \in \mathcal{A}(R')[n]$. Let $\sigma \in I_K$. Then $\mathcal{P}^{\sigma} - \mathcal{P} \in \mathcal{A}(R')[n]$ and

$$ \widetilde{\mathcal{P}^{\sigma} - \mathcal{P}} = \mathcal{O} $$

in $\tilde{\mathcal{A}}(k')$. Thus (since prime-to-p torsion injects into $\tilde{\mathcal{A}}(k')$ - by [1]) $\mathcal{P}^{\sigma} = \mathcal{P}$ and $P^{\sigma} = P$.