The question does not ask me to find the value of the game, merely to show that it's favorable. Being very lazy, I will do the minimum required.
As usual with such problems, it's convenient to consider a space of 5! equally likely outcomes, namely, all possible orderings of the 5 balls. (This means that I regard all balls as distinguishable, and I pretend that I keep drawing "for fun" after the game is officially over.)
Plainly, in any play of the game, I win a dollar, lose a dollar or break even. I just have to show that there are more winning than losing outcomes.
Clearly, any losing outcome ends with me drawing a gold ball on the fifth turn. Hence, the reversal of any losing outcome is a winning outcome, with me drawing gold on the first turn. This already shows that there are at least as many winning as losing outcomes.
To show that the game is favorable, all I have to do is find a winning outcome which is not the reversal of a losing outcome, e.g., an outcome of the form GSSSG whose reversal is also a winning outcome.
Question 1: It is easiest to prove this by induction. If $r = 0$ (there are no red balls) and $b \geq 1$, then clearly the first ball will be blue. That is, the number of balls $N$ that we need to draw in order to draw a blue ball has expected value $E(N) = 1$. Note that we can write this as
$$
E(N) = \frac{b+1}{b+1} = 1
$$
in accordance with the desired formula. Now, consider any other value of $r$, and $b \geq 1$ still. Suppose that we already have shown that for $r-1$ red balls and $b$ blue balls, the expected number of balls until we draw a blue ball is $\frac{b+r}{b+1}$, in accordance with the desired formula. We now draw a ball from an urn with $r$ red balls and $b$ blue balls. With probability $\frac{b}{b+r}$, the first ball is blue, and $N = 1$. With probability $\frac{r}{b+r}$, the first ball is red, and we now have an urn with $r-1$ red balls and $b$ blue balls. By the premise, we already know what the expected number of additional balls that need to be drawn in order to produce a blue ball; it is $\frac{b+r}{b+1}$. Therefore, for the urn with $r$ red balls and $b$ blue balls, we have
$$
\begin{align}
E(N) & = \frac{b}{b+r} \cdot 1
+ \frac{r}{b+r} \cdot \left(1+\frac{b+r}{b+1}\right) \\
& = \frac{b}{b+r} + \frac{r}{b+r} + \frac{r}{b+1} \\
& = 1 + \frac{r}{b+1} = \frac{b+r+1}{b+1}
\end{align}
$$
and we are done.
Question 2: By symmetry, the expected number of remaining balls left when only balls of one color (either red or blue) remain is equal to the number of balls of the same color we draw at the beginning. (To see this, draw all the balls out from the urn, and lay them out on a table, in sequence. The sequence from left to right is exactly as probable as the sequence from right to left.)
With probability $\frac{r}{b+r}$, the first ball is red. We are then left with $r-1$ red balls and $b$ blue balls, and we want to know how many balls we draw before (not until) we draw the first blue ball. This is one less than the result from Question 1. To this we have to add the first red ball. So when the first ball is red, our answer is
$$
N_{red} = \frac{b+r}{b+1}-1+1 = \frac{b+r}{b+1}
$$
With probability $\frac{b}{b+r}$, the first ball is blue, and by symmetry, the number of consecutive blue balls at the start is
$$
N_{blue} = \frac{b+r}{r+1}
$$
Thus, the expected number of consecutive balls drawn of the same color (which is also the answer to the original Question 2) is
$$
E(N) = \frac{r}{b+r} \cdot N_{red} + \frac{b}{b+r} \cdot N_{blue}
= \frac{r}{b+1} + \frac{b}{r+1}
$$
Best Answer
Your calculations on the third draw are off. Either you have selected two balls of the same color (probability $\frac{1}{3}$) in which case you are certain to win, or you have selected two balls of different color (probability $\frac{2}{3}$) in which case you win half of the time. We have:
First draw: $\frac{2}{4} \cdot 1 = \frac{1}{2}$
Second draw: $\frac{2}{3} \cdot 1 = \frac{2}{3}$
Third draw: $\frac{1}{3} \cdot 1 + \frac{2}{3} \cdot \frac{1}{2} \cdot 1 = \frac{2}{3}$
Fourth draw: $1$
Adding this all up, we arrive at $\frac{17}{6} \approx 2.83$.