To count the sequences of $r+b+g$ draws in which all of the red balls precede all of the green balls, note that they are obtained by starting with the sequence
$$\underbrace{RR\ldots RR}_r\underbrace{GG\ldots GG}_g$$
and inserting the $b$ blue balls arbitrarily into the sequence. This is a straightforward stars-and-bars problem; the link has both a formula and a pretty decent explanation of it, but if you have questions, leave a comment. Bear in mind that in any given sequence the $r$ red balls can actually be arranged in $r!$ orders, the blue balls in $b!$, and the green balls in $g!$, so that a given sequence of colors actually corresponds to $r!b!g!$ different sequences of balls drawn. However, since this is the same for each sequence of colors, it does no harm to count sequences of colors instead of sequences of balls.
Once you have that, you need only count the possible sequences of colors. You have to choose $r$ positions for the red balls, $b$ for the blue balls, ...
All of the boxes contain $N - 1$ balls. This is just a complicated conditional probability problem. Lets look at a single box with $r$ red balls and $g$ green balls. What would the probability be of getting green on the second? Well it depends on whether or not you draw a red or green first. If you draw a red first, then there are $\left.p(\text{second green } \right| \text{ first red}) = \frac{g}{r + g - 1}$. However, if you draw a green ball first then you have one less green to choose from giving: $\left.p(\text{second green } \right| \text{ first green}) = \frac{g - 1}{g + r - 1}$. So what are the chances of each condition happening? $p(\text{first red}) = \frac{r}{g + r}$ and $p(\text{first green}) = \frac{g}{r + g}$. Therefore we can finally write:
\begin{align}
p(\text{second green}) =& \left.p(\text{second green } \right| \text{ first red})p(\text{first red}) + \left.p(\text{second green } \right| \text{ first green})p(\text{first green})\\
=& \frac{r}{r + g}\frac{g}{r+g-1} + \frac{g}{r + g}\frac{g-1}{r+g-1} = \frac{g(r + g - 1)}{(r + g)(r + g - 1)} = \frac{g}{r + g}
\end{align}
Not surprising that drawing the second green has just as good of a chance of being green as the first pick.
Therefore for each of the $N$ boxes you need to compute $p(\text{second green})$ (which is just the probability of drawing a green on the first try). Now the condition is that we choose box $r$ which has $p(\text{second green}) = p(\text{first green}) = \frac{N - r}{N - 1}$. The probability of choosing box $r$ among $N$ boxes is just $\frac{1}{N}$ which gives:
$$
p(\text{second green}) = \sum_1^N \frac{1}{N}\frac{N - r}{N - 1} = \frac{1}{N(N - 1)}\sum_1^r (N - r)
$$
The first sum is very easy (you're just summing the same number, $N$, $N$ times) $\sum_1^N N = N\cdot N = N^2$. The second part is easy if you remember the sum of the first $n$ consecutive integers is $\sum_1^n i = \frac{n(n + 1)}{2}$. So this gives:
$$
p(\text{green}) = \frac{N^2 - \frac{N(N + 1)}{2}}{N(N - 1)} = \frac{2N^2 - N^2 - N}{2N(N - 1)} = \frac{N^2 - N}{2\left(N^2 - N\right)} = \frac{1}{2}
$$
For part $2$), we actually already computed that above: $\left.p(\text{second green }\right|\text{ first green}) = \frac{g - 1}{g + r - 1}$. But now you need to sum over the condition that it could be any of the $N$ boxes (edit: However, the last box, box $N$, has $0$ green balls (and thus seeing green first means it definitely wasn't this box. So we should only sum over the first $N - 1$ boxes and divide by $N - 1$, not $N$.):
\begin{align}
\left.p(\text{second green }\right|\text{ first green}) =& \sum_1^{N - 1}
\frac{1}{N - 1}\frac{N - r - 1}{N - 2} \\
=& \frac{N(N - 1) - (N - 1) - \frac{N(N - 1)}{2}}{N(N - 2)} \\
=& \frac{2N(N - 1) - 2(N - 1) - N(N - 1)}{2(N - 1)(N - 2)}\\
=& \frac{N(N - 1) - 2(N - 1))}{2(N - 1)(N - 2)} \\
=& \frac{(N - 1)(N - 2)}{2(N - 1)(N - 2)} \\
=& \frac{1}{2}
\end{align}
This is only valid for $N > 2$ (since if $N = 1$ there are no balls in each box and if $N = 2$ there is only one ball in each box). This result just confirms that drawing balls are independent events.
Best Answer
Looking at the final balls is the way to go, as you can regard drawing all $60$ balls as choosing one of the equally probable permutation. Working from the back, you can also ignore a colour once it has been drawn.
The probability that the last ball is blue and that the last green comes after the last red is $\dfrac{30}{10+20+30}\times \dfrac{20}{10+20} =\dfrac{1}{3}$ or more generally $b \times \dfrac{g}{1-b}$.
The probability that the last ball is green and that the last blue comes after the last red is $\dfrac{20}{10+20+30}\times \dfrac{30}{10+30} =\dfrac{1}{4}$ or more generally $g \times \dfrac{b}{1-g}$.
So the probability that all reds are drawn before the final blue and final green are drawn is $\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}$ or more generally $ \dfrac{bg}{1-b} + \dfrac{bg}{1-g}$.