The situation is easy to understand in the special case where $b=r$, since the whole situation is then symmetric between the two colors. Parts of the situation are asymmetric, for example the situation after you draw a blue ball on the first trial and add $c$ more blue balls. Obviously, blue is then favored. But there is an equally likely situation that equally favors red, so the overall situation is still symmetric.
Let me now try to explain a similar intuition in the non-symmetric case where $b\neq r$. Imagine that the $b+r$ balls initially in the urn are not only colored with two colors but also numbered, with $b+r$ distinct numbers, say numbers $1$ to $b$ for the blue balls and $b+1$ to $b+r$ for the red balls. Also, imagine that whenever a ball is drawn in a trial and is put back into the urn, the additional $c$ balls have not only the same color but also the same number as the ball that was drawn. (So, although the initial situation had different numbers on all the balls, later situations will have many balls with the same number.) Notice that, throughout the process, all balls numbered $1$ to $b$ will be blue and all balls numbered $b+1$ to $b+r$ will be red.
Now temporarily forget about colors and concentrate on numbers. The initial situation is symmetric between all $b+r$ numbers. Just as in the first paragraph above, the overall situation remains symmetric throughout the process. So all $b+r$ numbers have the same chance of being drawn in the $n$-th trial.
As before, there are asymmetries in conditional probabilities, for example the second ball drawn is more likely to have the same number as the first than to have any other particular number. But, as before, the asymmetries are equally likely to favor any of the numbers. So the overall probabilities are equal.
But (taking the colors into consideration again) this means that the probability of drawing a blue ball on the $n$th trial is $b/(b+r)$ because $b$ of the $b+r$ equally likely numbers correspond to blue balls.
We can use the reasoning behind linearity of expectation and indicator random variables.
Let $X$ be the number of white balls from our k draws.
$$X = X_1 + X_2 + X3 ... + X_k$$
where $X_i = 1$ if the $i^{\text {th}}$ ball selected is white and $X_i = 0$ if black.
$$E[X] = \sum_{i} E[X_i]= \sum_{i} P(X_i = 1)$$
$$P(X_i = 1) = \frac{n}{n + m}$$
$$E[X] = \sum_{i}P(X_i = 1) = \sum_{i = 1}^{k}{\frac{n}{n + m}}\\= \frac{kn}{n + m}$$
And I think a very similar reasoning can be used for the second part of your problem.
Best Answer
Hints: First, at the beginning of time $n+1$, there are exactly $n+2$ balls (why?). We then sample a ball: with probability $R_n/(n+2)$, we draw a red ball (why?), in which case $R_{n+1}=R_n+1$, while with probability $(n+2-R_n)/(n+2)$, we draw a white ball, so we are stuck with $R_{n+1}=R_n$. Combine these values and probabilities (how?) to get your answer.