Hi, this is only a partial answer to (3), that was too long to be a comment.
In a recent post of mine, Felipe Voloch pointed out a very useful tameness criterion proved by Gross ("A tameness criterion for galois representations..." Duke J. 61 (1990) on page 514).
In your case, $E$ is good ordinary at $p$ and the criterion for tameness applies. If your sequence $(\ast)$ splits, then $E(\mathbb{Q}_p)$ has a point of order $p$, and therefore $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ is diagonalizable (and, in fact, of the form $[\ast,0;0,1]$). It follows from Gross' criterion that $j(E)\equiv j_0 \bmod p^2$, where $j_0$ is the $j$-invariant of the canonical lift of the reduction of $E$.
Unfortunately, this is only a necessary condition, as $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ may be diagonalizable but without the lower right corner being trivial.
Theorem (originally due to Setzer?): Fix $E/\mathbb{Q}$ with $j(E)$ not 0 or 1728. Then for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ is isomorphic to $E[2](\mathbb{Q})$, so in particular $E_d(\mathbb{Q})_{tors}$ has order 1, 2, or 4. (Probably he also proved it for number fields.)
There's a paper of mine$^1$ with a much more general theorem using the theory of heights. I don't recall Setzer's proof except that it doesn't use heights.
Theorem: Let $K$ be a number field and let $A/K$ be an abelian variety with $\mu_n\subset {\rm Aut}(A)$. (This means we can twist $A$ by $n$'th roots of $d$.) Then every point $P\in A_d(K)$ satisfies one of the following two conditions:
- $P$ is fixed by a non-trivial $\zeta\in\mu_n$.
- $\hat h(P) \ge C_1(A)h^{(n)}(d) - C_2(A)$.
Here $\hat h$ is the canonical height relative to an ample symmetric divisor, and $h^{(n)}(d)$ is a sort of "$n$'th power free height," say equal to the minimum of $h(du^n)$ for $u\in K^*$. The constants depend on $A$, but are independent of $d$.
It follows from the theorem that after discarding finitely many $d \in K^*/{K^*}^n$, a point in $A_d(K)$ is either $1-\zeta$ torsion (hence $nP=O$), or its canonical height is positive, and hence it is nontorsion.
Of course, to describe more precisely what happens for the finitely many exceptional $d$ can be a delicate matter, as some of the other answers have indicated. I think it's interesting to see how one can approach the problem via heights or via representation theory.
$^1$ J.H. Silverman, Lower bounds for height functions, Duke Math. J. 51 (1984), 395-403.
EDIT: Fixed statement of first theorem. I'd originally written that "for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ has at most two elements." This is clearly false, since if $E$ has the form $E:y^2=(x-a)(x-b)(x-c)$ with $a,b,c\in\mathbb{Q}$, then $E_{d}[2](\mathbb{Q})$ has order 4 for every twist.
Best Answer
As has already been remarked, but more generally, if $K$ is a number field and $A/K$ and $B/K$ are abelian varieties that are isogenous over $K$, then the criterion of Neron-Ogg-Shafarevich has as a quick corollary that $A$ and $B$ have the same set of primes of bad reduction. The paper with the proof is by Serre and Tate. (Serre, Jean-Pierre; Tate, J., Good reduction of abelian varieties, Ann. Math. (2) 88, 492-517 (1968). ZBL0172.46101.) If you want a proof just for elliptic curves, you could look at Section VII.7 in (Silverman, Joseph H., The arithmetic of elliptic curves, Graduate Texts in Mathematics 106. New York, NY: Springer (ISBN 978-0-387-09493-9/hbk; 978-0-387-09494-6/ebook). xx, 513 p. (2009). ZBL1194.11005).