Here's a simple example where pev(A,BC)=pev(A,B)pev(A,C):

[EG1] P(A)=1/4, P(B)=P(C)=1/2, P(BC)=1/4, P(AC)=P(BC)=1/16, P(ABC)=1/64.

Then just suppose there are some medical diagnostics with these probabilities, etc. But see below for a comparison of this with your coin scenario.

More generally, let a=P(A)=1/4, b=P(B), c=P(C), x=P(BC), y=P(AC), z=P(AB), t=P(ABC).

The "before and after independence" assumption [BAI] is that x=bc and at=yz

The "independent tests" assumption [IT] is that (x-t)=(b-z)(c-y) and at=yz.

(2) How the "independent tests" assumptions are "stable"?

As I stated earlier, the condition A or the condition ~A screens off P(A|X).

Neither [BAI] nor [IT] is stable with respect to the variable a=P(A), and I see no equation involving anything I'd call "screening off", though there might be one somewhere.

In any case, that does not interest me, because your equation (7) has my attention:

(7) P(B|~AC) = (P(B) - P(A)(P(C|A)/P(C))P(B|A))/(P(~A)P(C|~A)/P(C))

A qualitative distinction between my would-be medical scenario [EG1] and your coin scenario is that medical diagnostics, and particularly their dependence/independence, are not always "causally justified", but two coin flips can be seen as independent because they visibly just don't interact.

I bet your (7) gives a good description of this somehow, or something closely related... But I still have to formalize my thoughts about this.

Let me think about it awhile, and I'll post again if I understand or wonder something more precise.