This sequence is the Thue-Morse sequence about which you can find quite a bit online; you can find a plethora of facts about it at the Online Encyclopedia of Integer Sequences as pointed out by Q the Platypus in the comments. (As a side note this is a really good website for finding sequences; you only need to type in the first 7 elements $0,1,1,0,1,0,0$ to it to get the Thue-Morse sequence). The Thue-Morse sequence has some interesting properties, including the fact that it describes a fair way to share as you noted.
The answer to your first question is yes, as Wolfram Mathworld confirms; specifically you can find here that:
[This sequence] is constructed by following a few simple steps:
(1) start with the two digit string, $01$
(2) replace every $0$ in the string by $01$, and replace every $1$ in the string by $10$.
(3) with the newly-created string from the previous step, go back to the beginning of step 2, and replace each $0$ and $1$ with the same values as before.
This (if you replace $0$ with $A$ and $1$ with $B$) is precisely your algorithm, which is an alternative way of generating the Thue-Morse sequence to the usual way which you mention (inverting the bits and adding the result on to the previous step - described here), so this substitution system does in fact generate the Thue-Morse sequence.
To answer part (a) of your second question, here you can find a mention of a way of calculating the $n^{th}$ term in the series:
To compute the nth element $t_{n}$, write the number n in binary. If the number of ones in this binary expansion is odd then $t_{n}=1$, if even then $t_{n}=0$.
This can be used to calculate who's turn it is if they know how many turns they have had, by simply taking the number $n$ of the turn they are up to, writing it in binary, and if there are an odd number of ones in the binary expansion then it is $B$'s turn and $A$'s turn if the number of ones is even. The same Mathworld page writes this in the simple form $t_n=s_{2}(n)\mod{2}$ where $s_{2}(n)$ is the binary digit sum.
As to part (b) of your second question, I do not know in general what information is required to be able to find your place in the series. However, simply knowing a past sequence of turns will not help if $A$ and $B$ do not know around how long they have been playing for, since if we let $S$ be the sequence of past turns that they do remember, and let $n$ be the actual position which they are at (which they want to find); then let $m=2^{\left\lceil\log_{2} n\right\rceil}$ be the next point at which the normal generating process will invert and add on the bits up to $m$.
However, the $m$ bits that follow bit $m$ are the inverse of the previous bits, and the $2m$ bits that follow bit $2m$ are the inverse of the first $2m$ bits. But this means that the sequence of bits from $3m$ to $4m$ are identical to the first $m$ bits, so the sequence $S$ occurs there also (since it occurs in the first $m$ bits). $S$ will also occur (as well as in many other places) between bits $15m$ and $16m$ and between bits $63m$ and $64m$ and so on. Thus $S$ actually occurs infinitely many times eventually in the Thue-Morse sequence, a property formally expressed here by noting that the Thue-Morse sequence is a uniformly recurrent word, i.e. for any past sequence $S$ they can remember, there is some length $l_S$ such that $S$ occurs in every block of length $l_S$ in the entire sequence (even though the Thue-Morse sequence is not periodic). The following image (where black dots are $1$s and white dots are $0$s) illustrates the recurrent nature of the sequence:
Thus if they know the last sequence of moves $S$, they would at the very least need to know around how many moves they have made already in order to be able to find where they are up to in the sequence, since the same sequence $S$ occurs infinitely many times throughout the game.
