Three comments: (a) Since the Galois closure $L$ of $K$ is a cubic extension of a quadratic imaginary field, maybe Heegner points will help. (b) magmas RationalPoints is quite efficient in finding points on elliptic curves over number fields, not sure if you have tried it. (c) I doubt this has anything to do with Duality's question that I answered because the curve does not have automorphisms of order 3. Twists don't help here. You could take the quadratic twist with respect to the quadratic subextension in $L$, but that doesn't help usually for finding points over $K$.

You can read off the ramification from the finitely many extension of $\mathbb{Q}_2$ that are left as option for the group $A_4$ and for the group $S_4$. It shouldn't be hard to check if any of them can occur as torsion extensions for some elliptic curve; I haven't done that.

$A_5$ is never the Galois group of an extension of $\mathbb{Q}_{\ell}$. The others can only occur for $\ell=2$.

Correction, it is at most 4. - But I should really stop thinking about this...

Oh, silly me, of course the rank is bounded. There is a 2-isogeny to the curve $y^2 = (x+5d)(x+2d)(x-7d)$. If $d$ is prime then only 2,3 and $d$ are bad, so from complete 2-descent, only taking the infinite local conditions, we know that the rank has to be at most 3. Probably one can get that down to 2 by looking at 2-adic local conditions. And a precise formula for the rank of the 2-Selmer group could be found and would depend on $d$ modulo some power of 2 and 3. But I won't have time to do this myself. Having to mark lots of 2-descents doen by my student in their exam.

This does not say anything about the rank itself. We expect as usual that 0 and 1 are the most frequent for each of the two root numbers. But I see no reason to think there aren't any with much larger rank.

Numerically, for $(d,6)=1$, it looks like the root number of the twist $E_d$ is equal to the root number at $2$ as $w_3=w_{\infty}=-1$ and $w_p=+1$ for $p\mid d$. It seems $w_2=-1$ if and only if $d\equiv 5 \pmod{8}$, but I haven't check this very far. Conjecturally on the finiteness of the Tate-Shafarevich group, we know the partiy conjecture, which would then give you a condition on the rank being even or odd only on $d$ modulo 8. --- But maybe I am completely wrong. It shouldn't be hard to do this explicitly as Joachim suggested already.

As this question came up again today, let me add that Coleman's conjecture has been proven in this article by Bullach, Burns, Daoud, Seo. Probably this answers this vague question.

I haven't checked. You will have to read the proof yourself, sorry.

The proof that torsion points have integral coordinates (at a place of good reduction) is basically the same as the injectivity. The proofs I know are either using the formal group directly or are inspired by that argument.

As the conductor is an effective divisor, $\operatorname{deg}(n_E)=0\iff n_E=0$.

It is the usual action of the quotient $G/H$ on $H^i(H,M)$ when $H$ is normal in $G$. To check that this is the same action as defined on $WC$ sounds like a good exercise to do. Be careful to view elements in the Weil-Chatelet groups as torsors, not just as curves.

The notation $\mathbb{Q}_p $ for a field other than the $p$-adics is confusing.