It turns out that you can do very slightly better than the dice in your example. The probability has to be a multiple of $1/36$, your example has $20/36$, and the best possible value is $21/36=7/12$.
From my answer to the question that anon linked to, we know that the probability of one $6$-sided die beating another while having a lower median can be at most $3/4-1/12=2/3=24/36$. Wikipedia gives an example of a cycle of four dice in which each die beats the next one with this probability.
You're looking for a cycle of three dice in which each die beats the next one with the same probability. Without loss of generality we can assume that the dice carry $18$ different numbers. Then there are $18!/6!^3=17153136$ different sets of dice, so it's well within the reach of our electronic friends to try them all out. Here's code that does that (rather inefficiently). The result is that the maximal probability for such a cycle is $21/36=7/12$ and there are eight different sets of dice that attain that probability. Here they are, with consecutive stretches of numbers collapsed into one as in your example:
0 3 3 3 6 6
2 2 2 5 5 5
1 4 4 4 4 4
0 0 3 3 3 6
2 2 2 2 2 5
1 1 1 4 4 4
0 3 5 5 5 5
2 4 4 4 4 7
1 1 1 6 6 6
0 3 5 7 7 7
2 2 4 4 6 9
1 1 1 8 8 8
0 3 3 5 5 5
2 2 2 4 4 7
1 1 1 6 6 6
0 3 5 5 7 7
2 2 2 4 6 9
1 1 1 8 8 8
0 3 3 3 3 5
2 2 2 2 4 7
1 1 1 6 6 6
0 0 3 6 6 6
2 2 2 5 5 5
1 1 1 4 7 7
There are two sets with $7$ different numbers, four sets with $8$ and two with $10$, and your near-optimal example has $9$, whereas the mean number of different numbers after collapsing, averaged over all sets with $18$ different numbers, is $18-17\cdot(5/17)=13$, so consecutive stretches of numbers and thus high collapsibility are a striking feature of these sets.
P.S.: I just realized that there's a symmetry in that you can invert the orders of both the dice and the numbers and get another set with the same probability. The first two sets are related by this symmetry, as are the third and seventh, the fourth and sixth and the fifth and eighth. Thus there are only four really different sets of dice, with $7$, $8$, $10$ and $8$ different numbers, respectively.
$a)$ Your answer is correct but you can get it easier. Out of the 36 cases, 6 cases are a tie, 15 G wins, 15 W wins (by symmetry). So the answer is $$\frac{15}{36}=\frac{5}{12}.$$
$b)$ The first roll is a tie with probability $\frac{1}{6}$. After that both have the same chance so the answer is $$\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}.$$
Another way of seeing this is to compute the probability that G wins in round $i$. We must have a tie in rounds $1,2,\dotsc,i-1$ and G winning in round $i$:
$$\left(\frac16\right)^{i-1}\frac5{12}.$$
Now we sum this up for $2\le i\le \infty$ to get the probability that G wins but not in round 1:
$$\sum_{i=2}^\infty \left(\frac16\right)^{i-1}\frac5{12}=\frac1{12}$$
$c)$ Suppose G gets a $2$. The probability of a tie is then $\frac{1}{6}$ and after that W wins with probability $\frac{1}{2}$. So the probability that W wins and the first round is a tie is $\frac{1}{12}$. The probability that W wins without a tie is $\frac{4}{6}=\frac{2}{3}$. So in total the probability of W winning is $$\frac{1}{12}+\frac{8}{12}=\frac{9}{12}=\frac{3}{4}.$$
Note that here we are looking for a conditional probability given that G gets a 2. This is already given as something that has happened. So we do not need to consider its probability in our calculations.
Best Answer
As you remark, for $n>6$ the probability that you land on $n$ is the average of the $6$ predecessor probabilities. As such, it can never exceed the maximum of those $6$.
It is clear that $n=6$ has the greatest probability of the first six values (easy to check this by hand, of course). Thus no value beyond $6$ can be more likely, as iterated averages must lower the maximum. Thus the maximum value occurs for $n=6$, for which the probability is just a little greater than $.36$
It's not even close. $P(5)\approx .30877$ and $P(n)<.3$ for all $n\neq 5,6$. The limiting value, for large $n$ is $\frac 1{3.5}\approx .28571429$ since the average toss of the die is $3.5$ This limit is reached fairly quickly, as $P(26)\approx .28574$