1) There is no algorithm that can compute arbitrarily many digits of $\Omega$. If we were in some fashion capable of checking whether given strings of numbers composed the first $n$ of $\Omega$'s binary expansion, we could simply brute force our way through all numbers with denominators $2^k,$ $k=1,2,3,\dots$, which would provide us with more and more digits of $\Omega$ - a contradiction.
However, all is not lost. First off, the set of rational numbers less than $\Omega$ is computably enumerable, so there exists an algorithm that lists them off one by one. More directly, we can compute successively better lower bounds by simply calculating known halting programs and discovering more of them (we never know how close we are, though, or otherwise we could extract digits). If the aliens give us a number less than Chaitin's constant, then in theory we could confirm that. In practice, though, we might well not have enough time and resources to check their number if it's too close to the real value, so we wouldn't be sure (and of course, we would be guaranteed failure if their number is $>\Omega$).
Second, just because no algorithm computes infinitely many digits, doesn't mean we can't soundly prove the first $n$ digits for some fixed $n$ (and a fortuitous choice of Turing machine). Mathworld provides two examples of Chaitin's constant being computed for two TM's:
$$\Omega_U= 0.0000001000000100000110\cdots_2$$
$$\Omega_U=0.0001000000010000101001110111000011111010\cdots_2 $$
For more on these, you'll have to see Calude (2002) and Calude, Dinneen (2007). Using these, we can do a rudimentary check on the alien's number by confirming or disconfirming that it agrees with what we have. But of course, this is extremely limited - and highly dependent on choice of Turing machine.
Third, the digits of $\Omega$ are algorithmically random, i.e. the first $n$ digits (alone) can only be enumerated in at least $n-O(1)$ steps. Thus it might be possible to investigate the plausibility of candidate digits via approximations of Kolmogorov complexity or other statistical tests. However, this may require a very large number of candidate digits in order to obtain a useful indicator, possibly more than is feasible to realistically work with.
2) Whether the answer to a general mathematics problem is "of any use" is outside the scope of computability theory. To the extent that proving or disproving a conjecture is inherently useful, then calculating truth or falsity based on $\Omega$ would be useful. If an existence claim turned out false, we could stop searching computationally, or if a universality claim turned out false we could start searching computationally for a counterexample. Precisely what the implications of a conjecture are depend on the problem itself, e.g. the Riemann hypothesis affects the distribution of prime numbers and subsequent arithmetic bounds in number theory.
However, a simple computation of this sort won't lend any insight into why a problem turns out the way it does. Normally in mathematics, famous unsolved problems require whole new theories, perspectives, or directions to resolve, but we won't get that from a $\Omega$-based calculation because all of the "reasoning" is not found in the digits, nor the checking of the digits. Chaitin's constant encodes only the answers, not the why's behind them.
Why do you think that $g$ is total recursive? That's a pretty huge leap.
I'm not sure why you think that defining $g$ as:
$$g(x) = f_x(x)+1$$
Is any different than:
$$g(x) = f_n(n)+1\text{ iff } k_n(x)=1$$
Since $k_n(x)=1$ exactly when $x=n$. The $k_i$ seem to be used to obfuscate. They don't make the definition of $g$ any more recursive, because $n\to f_n$ is still not recursive.
The only way to show that a function is recursive is to show a way to compute it. In particular, you can't use $f_x(x)$ to compute it, because the function $h(n,i)=f_n(i)$ is not recursive. Unless you can show a way to compute $g$ without reference to $f_n$ you have not show that $g$ is recursive.
Yes, if we could write infinite computer programs/Turing machines, or whatever, then your code for defining $g(0)=f_0(0)+1$, $g(1)=f_1(1)+1$, etc. would be allowed. But if you could write infinite programs, you could come up with any function $\mathbb N\to\mathbb N$, and your statement that there are only countably many recursive functions would be false.
Computability and recursive functions are about starting with finite definitions of functions. You have to completely tell me how to compute the function with a finite program. You have not done so.
Let's say that you have any function, $H:\mathbb N\to\mathbb N$.
By your reasoning, we could define:
$$G(x) = H(n)\text { iff } k_n(x)=1$$
Because we can define:
$$G(0) = H(0)$$
$$G(1) = H(1)$$
$$...$$
That's patently absurd - this is not a "method" for computing $G$, it's just a statement that $G=H$. If you don't know how to compute $H$, we certainly don't know how to compute $G$. But that's essentially your argument.
Best Answer
Note that the title question does not match the body. I'll answer the question in the body, since the title question is trivial.
Perhaps surprisingly, the answer is yes! That is:
Such an $f$ is called a Friedberg enumeration (or numbering); the existence of Friedberg enumerations was first proved by (unsurprisingly) Friedberg, but a simpler argument was later given by Kummer. I've written a bit more about these at CSSE and at MO.