The problem here is that Eq. 2 is not true. It is true that the change in potential energy is equal to the negative change of kinetic energy and viceversa, since if energy has to be conserved those two "energy containers" have to obey this. When you lose some potential energy it gets entirely transfered to the kinetic energy container.
So yeah, $\Delta V = -\Delta T$. But this does not mean that $V = -T$ at all. You could have one million joules stored as potential energy and the particle not moving at all ($100000\neq 0$ so $V\neq -T$) and a moment after you could still have $99000$ joules in potential energy and $1000$ in kinetic energy. You would indeed have $\Delta V = 99000-100000 = -1000 = -(1000) = - (1000-0) = - \Delta T$ but you wouldn't have in general that $V=-T$.
I assume that V+T=const is correct. Then if at the initial moment T=0 and the particle was far away from the center of gravity then V≈0. I see your point then..
That is indeed right. $V+T= constant$ is correct. And that's another reason why Eq. 2 is, in general, wrong. $V+T= constant$ means that $V = constant-T\neq -T$. The Hamiltonian would be $H = constant$ which makes sense since energy is indeed conserved.
It is also true that you can make that $contant=0$ since you are free to chose whatever zero value for the potential you want, and yes in that case you would indeed get Eq. 2 and have $H = 0$ which has nothing wrong about it and has perfect physical meaning if you have a well defined zero for the potential energy. But in general the answer should be $H = constant$ (assuming conservation of energy).