As I indicated on my retraction of my original statement in another post, I mistakenly thought you were another post member who consistently comes up with weak analysis.
With respect to your criticism about random variations: I have no choice but to deal with responses to my analysis by demonstrating ways for White to improve. To try to cover this opening in detail would require me to write a book on this thread.
As for the specifics: No one has disputed the fundamental attribute of my way of playing this opening i.e. White gets an active Queen that cannot be denied its role in stiffling the Black initiative. In no variation so far has anyone been able to demonstrate a way for Black to neutralize the active White Queen short of forcing her exchange with an early Qd5. As indicated on a previous post, this is not adequate for equality.
This is the positional/tactical way to shut down the Black initiative. Perhaps you could explain why a passive Queen in the 8.Be2 variation (With the White Queen stuck on either the first or second rank for a very long time) is more desirable than an active Queen in the 8.Qf3 variation.
I don't know as to whether bringing the queen out is a good idea. I am by no means an expert on the two knights' defense, nor do I have the latest theory on it.
But I think the original idea behind 8. Qf3!? was to try to make something of winning the c6-pawn as well, an idea that is now discredited and largely abandoned because it gives black a huge initiative.
Yes, the queen may be more active on f3/g3 than on d1, but she is also very vulnerable out there (getting pushed around), while black can hardly make a target of it while she's on d1. In addition you have to reckon with a possible attack on c2, which is now weak.
8.Qf3 Be7 9.Bxc6 Nxc6 10.Qxc6 Bd7 11.Qc4 0-0 12.Nc3 Rc8 13.Qe2
And now Black should play
13....h6! with suficient compensation
A) 14.Nge4 Nxe4 15.Qxe4 Bc6= (Vouldis-Sheram, 2006)
B) 14.Nf3 e4!
B1) 15.Ne4 Nxe4 16.Qxe4 Bb5! with (17...Re8 to follow) =/+ (Mas-Iuldachev, 2008)
B2) 15.Ne5, Be6!= 16.b3?,Rxc3 -/+ (Sacharov-Bakhmatov,