When trying to find the expectation of a "complicated" random variable, you can try using the very useful technique of writing the complicated variable as a sum of simpler variables and using the fact that the expectation of a sum of random variables is the sum of the individual expectations (note this is true of any sum, independence is not needed).
Here, for each $i=1$, $2$, $\ldots\,$, $n$, define the random variable
$$X_i=\cases{1,& if the $i^{\rm th}$ bin contains 2 or more balls,\cr0,&otherwise.}$$
Now let $X$ be the total number of bins that contain two or more balls. Note that
$X=\sum\limits_{i=1}^n X_i$.
Using the linearity of expectation:
$$
\Bbb E(X)=\sum_{i=1}^n \Bbb E(X_i).
$$
Note each $X_i$ is a Bernoulli variable, so $\Bbb E(X_i)=P[X_i=1]$ for each $i$. Moreover, this quantity is independent of $i$.
So, all you have left to do is to find the probability that a particular bin has two or more balls (here, I'd calculate the probability of the complement of this event and subtract from 1).
Suppose you throw balls into bins (without capping the bins) until twice the number of empty bins plus the number of bins with one ball is exactly equal to $2k-n$. Now, let's let $b_0$, $b_1$, and $b_2$ be the number of bins with exactly $0$, exactly $1$, and with $2$ or more balls, respectively.
We know that $$b_0 + b_1 + b_2 = k,$$
and $$ 2b_0 + b_1 = 2k-n.$$
Subtracting the second equation from twice the first one, we get that $$b_1 + 2b_2 = n.$$
Thus, if we throw balls into bins (without capping the bins) until the above condition is met, and remove all the excess balls from bins from two or more balls, the result is exactly the same as if we throw $n$ balls into $k$ bins, capping the size of the bins at $2$.
Now, let's look at expectations for a Poisson process. These won't directly prove your theorem, but you should be able to put good bounds on how much the expectation of a Poisson process differs from your exact ball-and-bin problem, and show that it will give you the corruct answer up to approximately $1/\sqrt{k}$.
A Poisson process with rate $\lambda$ will yield an expected number of $e^{-\lambda} k$ bins with $0$ balls, and $\lambda e^{-\lambda} k$ bins with $1$ ball. Thus, to find the rate corresponding to $n$ balls, we need to solve the equations:
$$ 2 e^{-\lambda} k + \lambda e^{-\lambda} k = 2k - n,$$
or
$$ 2 - e^{-\lambda} (2 + \lambda) = \frac{n}{k}.$$
We can solve this equation numerically and plot the fraction of bins that are filled as the ratio of balls to bins, $n/k$, goes from $0$ to $2$.
If the number of bins equals the number of balls, roughly 31.8% of the bins are filled with two balls. If there are three balls for every two bins, roughly 61.8% of the bins are filled.
In fact, the same solution will apply to any maximum capacity, although solving the equations will get harder.
Best Answer
You want to use what are called indicator variables. To see how this works, let's take your first problem. Let $Y$ be the number of bins that are empty. You want $E[Y]$. Now define the indicator variables $X_i$ so that $X_i$ is $1$ if bin $i$ is empty and $0$ otherwise. Then we have
$$Y = X_1 + X_2 + \cdots + X_n.$$
By linearity of expectation,
$$E[Y] = E[X_1] + E[X_2] + \cdots + E[X_n].$$
So now the problem reduces to calculating the $E[X_i]$ for each $i$. But this is fairly easy, as $$E[X_i] = 1 P(X_i = 1) + 0 P(X_i = 0) = P(X_i = 1) = P(\text{bin $i$ is empty}) = \left(1 - \frac{1}{n}\right)^m,$$ where the last equality is because balls $1, 2, \ldots, m$ must all go in a bin other than $i$, each with probability $1 - \frac{1}{n}$.
Therefore, $$E[Y] = n \left(1 - \frac{1}{n}\right)^m.$$
Your second problem can be worked in a similar fashion. Let $Y$ be the number of bins that have exactly $1$ ball. Let $X_i$ be $1$ if bin $i$ has exactly one ball and $0$ otherwise. Then all that's left is to figure out $E[X_i]$, which is the probability that a given bin has exactly $1$ ball, and go from there. Since this is homework, I'll let you finish.