The probability is exactly $\frac{1}2$ in the original game. The crucial fact is that rolling a $5$ is a loss and that rolling a $6$ is a win - so we can consider the game over once one of those appears. However, at any given roll, the probability of rolling a $5$ equals the probability of rolling a $6$. We can quickly extend this local property to learn that the game is fair:
Suppose player $X$ won, rolling a $6$. An equally likely outcome would be that the game progressed the same until the last roll, at which point $X$ rolled a $5$. Similarly, if $X$ lost rolling a $5$, it was equally likely that they'd have rolled a $6$ instead.
That is, we can pair up all the possible games by switching the last roll between $5$ and $6$, which changes who won, but not the probability of such events occurring.
Extension to Bonus (b)
Notably, if we stop the game whenever we reach a value of $k$ or more, then it is clear that the probability that we stop the game at a value of exactly $k$ is $\frac{1}n$, where $n=7-k$ is the number of integers in $[k,6]$ - that is, the number of rolls greater than or equal to $k$. This holds because switching the last value in such a game has similar effect to before, and hence the last value is uniformly distributed. In particular, this means that the probability of $k$ appearing in a normal game (since the game after a value of $k$ or more is irrelevant as $k$ will not appear again) is $\frac{1}n$.
We can rephrase the above as: We expect to see $\frac{1}n$ turns where $k$ is passed to the other player. Then, the expected length of a game is the sum of the expected number of turns wherein each given value is passed due to the linearity of expectation - thus, the expected length of the game is
$$1+\frac{1}2+\frac{1}3+\frac{1}4+\frac{1}5+\frac{1}6=\frac{49}{20}$$
The number of rolls is obviously distributed the same as if we just rolled a die until we got a $6$. We can see that the expected number of rolls $r$ satisfies the relation
$$r=\frac{1}6\cdot 1 + \frac{5}6(r+1)$$
since we take $1$ roll $\frac{1}6$ of the time, and otherwise expect $r+1$ rolls. This solves to
$$r=6.$$
More generally, we expect as many rolls as there are faces on the die.
Alternate Bonus (b) Answer
Suppose that, in the previous turn, a player passed the number $6-n$. Let the expected number of turns remaining be $t_n$. Then, we have $t_0=0$ and $t_6$ is the expected number of turns in a game with a six sided die. It can be seen that $t$ satisfies the relation:
$$t_n=1+\frac{1}n\sum_{i=0}^{n-1}t_i$$
That is to say, each $t_i$ is one more than the average of those below it. This looks unwieldily, but it actually turns out to equal the well-known harmonic series. In particular, define
$$T_n=\frac{1}{n+1}\sum_{i=0}^{n}t_i$$
as the average of the sequence thus far. We can express this algebraically, but consider the following:
The average of the sequence $t_0,t_1,\ldots,t_{n-1}$ is $T_{n-1}$. Thus, so is the average of $t_0,t_1,\ldots,t_{n-1},T_{n-1}$, as putting the mean into the sequence doesn't affect the mean. Increasing one element of the $\frac{1}{n+1}$ elements by $1$ increases the mean by $\frac{1}{n+1}$. Therefore $T_n=\frac{1}{n+1}+T_{n-1}$.
From this, and the fact that $t_n=T_{n-1}+1$, we can use the relation that $T_{n-1}=\frac{1}n+T_{n-2}$ as derived above to yield:
$$t_n=\frac{1}n+T_{n-2}+1$$
$$t_n=\frac{1}n+\frac{1}{n-1}+T_{n-3}+1$$
and expanding until $T_0=0$ yields
$$t_n=\frac{1}n+\frac{1}{n-1}+\ldots+1$$
meaning that $t_n$ is the $n^{th}$ harmonic number - the sum of the reciprocals of the first $n$ natural numbers. So, this game's expected length, which is $t_6$ is
$$\frac{1}1+\frac{1}2+\frac{1}3+\frac{1}4+\frac{1}5+\frac{1}6=\frac{49}{20}$$
Though there's no particularly helpful closed form for exact computation of Harmonic numbers as ratios of two integers, this means that, when played with an $n$ sided die, the game will take about $\log(n)+\gamma$ turns where $\gamma$ is the Euler-Macaroni constant.
