Not sure what you did wrong, but I suspect that you ignored something in your casework. Also, the total number of cases should be $n^{n+1}$, not $n^{n+2}$. Here is an example of a much simpler approach.
First, calculate the number of ways there can be two empty bins. The number of ways for the two empty bins to be chosen is
$$_{n}C_2=\frac{n(n-1)}{2}$$
then, since there must be exactly two empty bins, we must place one ball in each of the remaining $n-1$ bins so that none of them are empty. We then have one last ball with $n-1$ choices. Thus the number of ways to place the balls and have exactly two empty bins is
$$\frac{n(n-1)}{2}*(n-1)$$
$$\frac{n(n-1)^2}{2}$$
And so the probability is
$$\frac{\frac{n(n-1)^2}{2}}{n^{n+1}}$$
$$\frac{(n-1)^2}{2n^{n}}$$
EDIT: This is incorrect. By saying that the number of ways to distribute the balls is $n^{n+1}$, I imply that the balls are distinct objects; however, if they are distinct, then the number of ways to place one ball in each bin other than the two designated empty bins is $(n-1)!$, and so the number of ways to place the balls and have two empty bins is
$$\frac{n(n-1)}{2}*(n-1)!*(n-1)=\frac{n(n-1)^2(n-1)!}{2}$$
and so the probability is
$$\frac{\frac{n(n-1)^2(n-1)!}{2}}{n^{n+1}}$$
$$\frac{(n-1)^2(n-1)!}{2n^n}$$
The placement of the non-red balls is irrelevant...the problem is decided by the placement of the red ones alone.
To do the first part, it is easier to work from the complimentary event. The probability that a specified red ball goes to a bin other than the second is $\frac 34$. Thus the answer is $$1-\left( \frac 34\right)^2=\boxed {\frac 7{16}}$$
To do the second one, it might be helpful to simply list the possible placements of the two red balls. Given that the second bin must contain a red ball, there are only four cases: $$(1,1,0,0)\quad (0,2,0,0)\quad (0,1,1,0)\quad (0,1,0,1)$$
Where writing, say, $(1,1,0,0)$ means that the first two bins each have one red ball and the second two each have none.
Note that the probability of getting $(0,2,0,0)$ is $\frac 1{16}$ since we must have both red balls going to the second bin. The probability of getting, say, $(1,1,0,0)$ is $\frac 18$ since we can get this configuration in two ways (either the first red ball goes in the first bin and the second red ball goes in the second bin, or the first red ball goes in the second bin and the second red ball goes in the first bin). This gives another way to see $\frac 7{16}$ as the result for the first problem, since $$\frac 18+\frac 1{16}+\frac 18+\frac 18=\boxed {\frac 7{16}}$$
We see from this that the conditional probability that the first bin has a red ball (conditioned on the fact that the second bin also has one) is $$\frac {1/8}{7/16}=\boxed {\frac 27}$$
Side note: if you prefer to work with equi-probable scenarios (not a bad idea) then you need to indicate the placement of $r_1, r_2$ (the two red balls) separately. Thus the scenario $(1,1,0,0)$, say, becomes two scenarios, $(r_1,r_2,0,0)$ and $(r_2,r_1,0,0)$. You wind up with seven scenarios in which the second bin contains at least one red ball, each of which having probability $\frac 1{16}$.
Best Answer
One can easily use the complement-rule here. Let $A$ denote the set of all possible arrangements. Let $A_1$ and $A_2$ $A_{1,2}$ denote the events that (respectively) bin 1 and 2 are empty. It should be clear that: $A= A_1 \cup A_2 \cup (A_1 \cup A_2)^c.$
It then follows that (because $A_1$, $A_2$ and $(A_1 \cup A_2)^c$ are disjunct and finitely countable):
$A= \vert A_1 \vert + \vert A_2\vert + \vert(A_1 \cup A_2)^c\vert$.
So it simply says, that the number of configurations without empty bins is all the configurations minus the (two) possible configurations with one empty bin (as it is impossible to have them both empty).