For part (i) I solution I have tried the following. Let $N$ be the number of balls drawn before and including the last white ball drawn. Using the hyper geometric distribution:

\[ P(N \leq n) = \frac{\binom{w}{w} \binom{b}{n-w}}{\binom{b+w}{n}} \]

Which simplifies to:

\[ P(N \leq n) = \frac{b! \times n!}{(b+w)! \times (n-w)!} \]

Also
\[ P(N = n) = P(N \leq n) - P(N \leq n-1) \]
\[ \therefore P(N = n) = \frac{b! \times n!}{(b+w)! \times (n-w)!} - \frac{b! \times (n-1)!}{(b+w)! \times (n-1-w)!} \]

Which simplifies to:
\[ P(N = n) = \frac{b! \times (n-1)! \times w}{(b+w)! \times (n-w)!} \]
for $w \leq n \leq b + w$ and 0 otherwise
\[ \therefore E(N) = \sum^{b+w}_{n = w} \frac{b! \times (n-1)! \times w}{(b+w)! \times (n-w)!} \times n \ = \sum^{b+w}_{n = w} \frac{b! \times n! \times w}{(b+w)! \times (n-w)!} \]

At which point I hit a brick wall, assuming what I've said to the point isn't nonsense. I've also had a look on TSR which gives a solution based on the binomial distribution. To quote:

"We may consider lining the balls up in a row and asking for the expected number of black balls to the left of the rightmost white ball. For each black ball, the probability that it is placed to the left of the last white ball is $\frac{w}{w+1}$ so the number of black balls to the left of the last white ball has a Binomial distribution with $n=b$ and $p=\frac{w}{w+1}$ so the expected number is $\frac{bw}{w+1}$"

I don't know how the probability of $p=\frac{w}{w+1}$ arises or why the binomial distribution would be appropriate. I would be grateful for any clarification.