Notice that if the formula on $P(B_m \cap B_n)$, we would completely determine $P(B_m \cap R_n)$ since $P(B_m \cap B_n) + P(B_m \cap R_n)=P(B_m) $.
\begin{align}
&P(B_m \cap B_{m+k+1})\\&=P(B_{m+k+1} | R_{m+k}, B_m)P(R_{m+k}|B_m)P(B_m)
\\&+P(B_{m+k+1} | B_{m+k}, B_m)P(B_{m+k}|B_m)P(B_m)
\end{align}
Let's compute the individual terms.
$$P(B_m) = \frac{b}{b+r}.$$
Also, by induction hypothesis,
$$P(R_{m+k}|B_m)=\frac{P(R_{m+k} \cap B_m)}{P(B_m)} = \frac{\frac{br}{(b+r)(b+r+c)}}{\frac{b}{b+r}}=\frac{r}{b+r+c}.$$
We drew a black ball at round $m$, also, we know that at round $m+k$, there are a total of $b+r+(m+k-1)c$ balls, of which $\frac{r}{b+r+c}\cdot [b+r+(m+k-1)c]$ of them are red and hence $\frac{b+c}{b+r+c}\cdot [b+r+(m+k-1)c]$ of them are blue.
At round $m+k+1$, there are a total of $b+r+(m+k)c$ balls, the number of blue ball didn't increase if we previously draw a red ball. Hence, we have
$$P(B_{m+k+1}|R_{m+k}, B_m )=\frac{\left( \frac{b+c}{b+r+c}\right) [b+r+(m+k-1)c]}{b+r+(m+k)c}$$
We have $$P(B_{m+k}|B_m)=\frac{b+c}{b+r+c}$$
We drew a black ball at round $m$, also, we know that at round $m+k$, there are a total of $b+r+(m+k-1)c$ balls, of which $\frac{r}{b+r+c}\cdot [b+r+(m+k-1)c]$ of them are red and hence $\frac{b+c}{b+r+c}\cdot [b+r+(m+k-1)c]$ of them are blue.
At round $m+k+1$, there are a total of $b+r+(m+k)c$ balls, the number of blue ball increased by $c$ if we previously draw a red ball. Hence, we have
$$P(B_{m+k+1}|B_{m+k}, B_m )=\frac{\left( \frac{b+c}{b+r+c}\right) [b+r+(m+k-1)c]+c}{b+r+(m+k)c}$$
Now, we have each individual term and we just have to substitute them back to compute $P(B_m \cap B_{m+k+1})$.
\begin{align}
&P(B_m \cap B_{m+k+1})\\
&=\frac{\left( \frac{b+c}{b+r+c}\right) [b+r+(m+k-1)c]}{b+r+(m+k)c} \cdot \frac{r}{b+r+c} \cdot \frac{b}{b+r}\\
&+ \frac{\left( \frac{b+c}{b+r+c}\right) [b+r+(m+k-1)c]+c}{b+r+(m+k)c} \cdot \frac{b+c}{b+r+c} \cdot \frac{b}{b+r}\\
&= \left(\frac{b(b+c)}{(b+r)(b+r+c)^2(b+r+(m+k)c)} \right)\cdot\\& \left((b+r+(m+k-1)c )r+(b+r+(m+k-1)c )(b+c) + c(b+r+c)\right) \\
&= \left(\frac{b(b+c)}{(b+r)(b+r+c)(b+r+(m+k)c)} \right)\cdot \left(b+r+(m+k-1)c + c\right) \\
&= \frac{b(b+c)}{(b+r)(b+r+c)}
\end{align}
Best Answer
There are two cases
In first draw ball is red and again it's red , probability to it is $$\frac{r}{r+b}.\frac{r}{r+b}$$
In first draw ball is black and again it's red , probability to it is $$\frac{b}{r+b}.\frac{r}{r+b+2}$$
so required probabilty is $$P(R)= \frac{r}{r+b}.\frac{r}{r+b}+\frac{b}{r+b}.\frac{r}{r+b+2}$$
Similarly
$$P(B)=\frac{r}{r+b}.\frac{b}{r+b}+\frac{b}{r+b}.\frac{b+2}{r+b+2}$$