Here is another way answer to (a). (And I have posted about this before on MO here, here, here and here.)
The idea is to consider the Turing machine programs as having random instructions. A single line in a Turing machine program is an instruction of the form: when in such-and-such a state, reading such-and-such a symbol, then write such-and-such, move in such-and-such a direction and change to such-and-such a new state. We may consider the collection of all programs having at most $n$ states, and for any fixed $n$, one may consider the notion of a random $n$-state program; every line of the program is chosen randomly, with uniform probability. The question becomes, then, what is the resulting typical nature of a random program?
For any desired behavior, one may consider the set $A$ of programs exhibiting that behavior. What we want is some way to measure the size of $A$. It seems quite natural to use the asymptotic density of the members of $A$ among the $n$-state programs, as $n$ becomes large. That is, the probability that a Turing machine program has property $A$ is the limit $$\lim_{n\to\infty} \frac{|A\cap P_n|}{|P_n|},$$ if this limit exists, where $P_n$ is the collection of all $n$-state Turing machines. For example, if the density is $1$, it means that more than $99\%$ of the $n$-state Turing machine programs are in $A$, as $n$ becomes large, and more than $99.9\%$, as close to $1$ as desired.
Your question was, what is the probability that a program halts? The answer, if one uses one of the standard Turing machine models, with a one-way infinite tape and a finite alphabet, and a single halt state, is that the probability is zero. This follows from the main argument of my paper
What we prove in that paper is that there is a set $A$ of Turing machine programs which consists of almost all Turing machine programs, in the sense that the asymptotic density of $A$ is $1$, such that membership in $A$ is linear-time decidable and also the halting problem for programs in $A$ is linear time decidable. Thus, the halting problem is decidable with probability one. This is an instance of the black-hole phenomenon, by which the difficulty on an infeasible or intractible problem is concentrated in a very small region, of measure zero, outside of which the problem is easy. Our main point is that even the halting problem admits a black hole. Clearly it will not do to base an encryption scheme on the difficulty of a problem with a black hole, since if the criminals can rob the bank $95\%$ of the time, or even $5\%$ of the time, it is bad enough.
The way the argument proceeds is by showing something more, namely, that for the standard model of computability I mentioned above, the probability one behavior of a Turing machine is that the head falls off the tape before a state is repeated. Thus, if this behavior is regarding as non-halting, than almost all programs do not halt on a fixed input. (But if having the head falling off the tape is regarded as halting, then almost all programs halt; the main point is that the behavior of a random program for this standard model is trivial in this way.) The proof appeals to the Polya recurrence phenemenon, with the basic argumment being that if every new line of a program is random, then as long as the state has no yet repeated, then the odds of moving left or right make the behavior like a random walk, and so with probability one, the head will return to the left-most cell and fall off the tape.
Unfortunately, the argument depends on the computational model, and the question is open for the two-way infinite tape model.
