For (i) I think you need to consider the players winning after $k$ rounds with $k=1,2,3,...$
Note that for player 1 to win in $k$ rounds, it means that he lost his first $k-1$ rounds, player 2 lost his first $k-1$ rounds, and player 1 won his $k$th round. By independence, this is ($A_k=$"player 1 wins after k rounds", $B_k$ is defined similarly for player 2):
$$P(A_k)=(4/6)^{k-1}(2/6)^{k-1}(4/6)=(2/9)^{k-1}(2/3)$$
Hence $$P(\text{1 wins})=\bigcup_{k=1}^\infty A_k=\sum_{k=1}^\infty P(A_k)=(2/3)\sum_{k=1}^\infty (2/9)^{k-1}={2 \over 3}{1 \over 1- {2 \over 9}}$$
Now for $B_k$ to occur player 1 must loose his first $k$ rounds, player $2$ his first $k-1$ and player 2 must win his $k$th round, which gives:
$$P(B_k)=(4/6)^k(2/6)^{k-1}(4/6)$$
$$P(\text{2 wins})={4 \over 9}\sum_{k=1}^\infty (2/9)^{k-1}={4 \over 9}{1 \over 1- {2 \over 9}}$$
It follows that $P(A)>P(B)$
ii) is fine.
For iii) again the probability of a game of length k won by 1 is $(2/9)^{k-1}(2/3)$ and won by 2 is $(4/9)(2/9)^{k-1}$. Hence the longer games are less probable, and since $2/3>4/9$ the most probable run is 1 winning is 1 round, hence a game of length 1.
Yes, it's a correct approach. Here's another, which you might find simpler.
Let $p$ be the probability that Obediah wins the game. She has a probability of $\frac 16$ of winning the game on her first roll. If she doesn't, then the players have switched positions so that Nathaniel now has a probability of $p$ of winning the game.
Thus, $p+ \frac 56 p =1$, so $p = \frac {6}{11}$ and Nathaniel's probability of winning is $1-p=\frac{5}{11}.$
Best Answer
This game has two states.
Starting: The total is 0, and whatever the first player rolls, the turn passes to the next player and the state changes to standard.
Standard: The total is greater than 0, but not a multiple of 7. The player rolling has a 1/6 chance to win; otherwise, the turn passes to the other player.
Does that make the strategy clear?