I do not know whether the more qualified members here consider this question suitable for this site, but anyway here is a partial answer to your question (this a bit too long for a comment anyway). I think the set $E_2$ must contain an infinite number of elements.

Here is my reasoning. Let's first notice that the configurations you have written as $T_x$ correspond to giving an input $x-1$ to the machine. The first thing we can observe is that given a program with access to the function $F$ (defined in your original question) one can calculate the characteristic set corresponding to the set TOTAL. TOTAL being the set containing indexes of all (ordinary) programs that halt on every input. Similarly, given a program with access to the characteristic function of TOTAL one can compute the function $F$ (just observe that $H \le_T TOTAL$). Both of these reductions are relatively clear.

Now, it seems to me, the question we should ask is whether the function $F$ is $H$-computably bounded, where $H$ is the halt set. I think the answer should be in negative. Here is the reason. Suppose, by contradiction, that $F$ is indeed $H$-computably bounded by some function $F_B$. Then we can actually show that characteristic function of TOTAL is $H$-computable, or in other words $TOTAL \le_T H$ (which is indeed well-known to be false). To see this, suppose we have an enumeration of all ordinary programs (without access to any oracale). Consider any program with index $i$. We can simply use the halt oracle to check the halting of the given program for all inputs from $0$ to $F_B(i)$. Here $F_B$ is a function such that $F_B(x) > F(x)$ for all $x$. If the program loops on even just one of the given inputs, then we return 0/false (meaning the program with index $i$ isn't total). If the program halts on all inputs in given range from $0$ to $F_B(i)$, then we can actually return 1/true (meaning the program with index $i$ is total). The last sentence is basically true because if the program ever looped on any input it would definitely have looped on input $F(i)-1$ with $F(i) \ge 1$, by definition of $F$. Yet we checked the values till $F_B(i)$, where $F_B(i) > F(i)$.  

Now, as to the question of why this means $E_2$ contains an infinite number of elements, I haven't thought it over rigorously, but the numerator in the fraction given in question is a function which isn't $H$-computably bounded while the denominator is a function which is $H$-computably bounded. So by analogy (assuming $f$ to be a function which isn't computably bounded) if I compare with a fraction such as $\frac{f(x)}{x^2}$, it seems 'obvious' to me that definining a set such as $\{\frac{f(1)}{1^2},\frac{f(2)}{2^2},\frac{f(3)}{3^2},\frac{f(4)}{4^2},.....\}$ will contain an infinite number of elements (I have assumed rounding to lower integer after division). This should be made more rigorous though.