Case 1
You picked a white ball from the first box.
This happens with a chance of $3/5$. Putting it into the second box, now there are $5$ white and $4$ black balls. Now the chance to pick a white ball is $5/9$. So the total chance in case 1 is $3/5 \cdot 5/9 = 1/3$.
Case 2
You picked a black ball from the first box.
This happens with a chance of $2/5$. Putting it into the second box, now there are $4$ white and $5$ black balls. Now the chance to pick a white ball is $4/9$. So The total chance in case 2 is $2/5 \cdot 4/9 = 8/45$.
Combining the two cases, the chance is $$1/3 + 8/45 = 23/45 \approx 51\%.$$
The first box has $a$ white and $b$ black, the second box has $b$ white and $a$ black. One ball is either moved from the first box to the second (with probability $1/2$) or from the second box to the first (also probability $1/2$). You then select a ball from the box to which a ball was added. You are looking for the probability that the box you select from is the first box, given that it is white.
Let $A$ be the event that the box you select from is the first box. Since both boxes are equally likely to be the box that is added to, $P(A) = 1/2$.
Let $B$ be the event that the ball you select is white. By symmetry, $P(B) = 1/2$.
You're looking for $P(A \mid B$), so you apply Bayes' theorem: $$P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}$$
Since $P(A) = P(B) = 1/2$, this means $$P(A \mid B) = P(B \mid A)$$
$P(B \mid A)$, the probability that the ball is white, given that you choose from the first box, is considerably easier to calculate. The fact that you choose from the first box means that a ball was taken from the second box and put into the first.
The probability $P(W)$ that a white ball was taken from the second box and added to the first is $$P(W) = \frac{b}{a+b}$$
The probability $P(B)$ that a black ball was taken from the second box and added to the first is $$P(B) = 1 - P(W) = \frac{a}{a+b}$$
Now, we write $P(B \mid A)$ in terms of $P(W)$, $P(B)$, $a$, and $b$:
$$P(B \mid A) = P(W) \cdot \frac{a+1}{a+b+1} + P(B) \cdot \frac{a}{a+b+1}$$
The terms multiplied by $P(W)$ and $P(B)$ are the probabilities of getting a white ball, given a white-ball-transfer and a black-ball-transfer, respectively. In each case, this is just the number of white balls in the first box, divided by the total number of balls in the first box.
$$P(B \mid A) = \frac{b}{a+b} \cdot \frac{a+1}{a+b+1} + \frac{a}{a+b} \cdot \frac{a}{a+b+1}$$
$$P(B \mid A) = \frac{a^2 + ab + b}{a^2 + 2ab + b^2 + a + b}$$
Finally,
$$P(A \mid B) = \frac{a^2 + ab + b}{a^2 + 2ab + b^2 + a + b}$$
Best Answer
Think of it visually. First you have the configuration
In the first draw we pick a ball from box #1. Since it is equally likely to pick any of the balls we have 7 of 12 possible balls to pick from. We therefore get $$ P(\text{white from #1 to #2}) = \frac{7}{12} $$ We now assume that we succeded in picking a white ball. We therefore base our next probability on the condition that we had the event "white from #1 to #2" in the first draw.
Under this condition our configuration looks like this
If we have a condition prevalent we always write that as conditional probability, that is what the bar in the other probability term means. We now draw two times from box #2 and each time we want to draw a red ball. Just as above we have 10 good and 19 total cases for the first draw from box #2 and 9 good and 18 total cases in the second. (This is only the case if we have moved a white ball to box #2 in the very first draw, that is why we use conditional probability)
To get the probability we multiply the probabilities for the draws from box #2. We therefore get $$ P(\text{rr from #2 to #1 | w from #1 to #2}) = \frac{10}{19}\cdot\frac{9}{18} = \frac{5}{19} $$ All in all we now get $$ P(\text{7 red balls in box #1}) = \frac{7}{12}\cdot\frac{5}{19} = \frac{35}{228}. $$