Let $X$ denote the number of white balls selected when $k$ balls are chosen randomly from an urn containing $n$ white balls and $m$ blacks.
For $i＝1....k; j＝1....n$,
$$\begin{align*} X_i&＝\begin{cases} 1,&\text{if }i\text{-th ball selected is white}\\ 0,&\text{otherwise} \end{cases}\\ Y_j&＝\begin{cases} 1,&\text{if white ball }j\text{ is selected}\\ 0,&\text{otherwise} \end{cases} \end{align*}$$
Compute $E[X]$ in two ways by expressing $X$ first as a function of the $X_i$s and then of the $Y_j$s.
Sorry for the poor codes
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.