For $a$ the point is that $Prob[A_j] \le nProb[A_j^1]$ because each slot has the same chance to have at least $j$ entries. It will actually be less than this because the right side counts cases where two slots each have $j$ entries twice while the left side counts them once. We were asked for an upper bound, so this is not a problem.
For $b$ the idea is to assume that each key maps to a slot randomly, so the chance a given key maps to slot $1$ is $\frac 1n$
For $c$ you just use $b$ and multiply by the number of subsets of size $j$
Let $\mathbf{1}_{ij}$ denote the indicator random variable for the event that keys $i$ and $j$ are mapped to the same bucket. You are then looking for $\mathbb{E}\left[ \sum_{1\leq i\leq j \leq n} \mathbf{1}_{ij} \right].$ By linearity of expectation, that is
$$ \sum_{1\leq i \leq j \leq n} \mathbb{E}[\mathbf{1}_{ij} ]$$
There are $\binom{n}{2} = \frac{n(n-1)}{2}$ such $i,j$ pairs that we are summing over. Each term is $$\mathbb{E}[\mathbf{1}_{ij}] = \mathbb{P}(\text{keys i and j are mapped to the same bucket}).$$
Since they are mapped uniformly and independently into $m$ buckets, the probability that two keys are mapped to the same bucket is $\frac{1}{m}.$ So the 3rd option is the correct answer.
The step going from "there are $n/m$ expected elements in any bucket" to "so the number of colliding pairs is.." does not work. For example, $n/m$ may not be an integer. Using this approach, we would need to figure out the distribution of the bucket counts. Say we have $2$ buckets and $2$ keys. The two buckets can have the following number of keys in them: $(0, 2), (0,2)$ and $(1,1). (1,1)$ occurs with probability $1/2$, the others have probability $1/4$. Then for each case, we apply your logic to get the number of pair collisions. For $(0,2)$, there's $0$ pairs colliding in bucket $1$ and $1$ pair in bucket $2$. So in this case, there is $1$ colliding pair and this case happens with probability $1/4.$ In the $(1,1)$ case there are $0$ colliding pairs in total, and this case has probability $1/2.$ In the $(0,2)$ case we again have $1$ colliding pair, and that case is probability $1/4.$ So the expected value for $n=2, m=2$ is $$1\cdot(1/4) + 0\cdot(1/2) + 1\cdot(1/4) = 1/2. $$
In principle we can do this same computation for general $n,m$ but these are the types of questions where linearity of expectation comes in so useful - when we only want an expected value, linearity of expectation often allows us to dodge actually computing the distribution.
Best Answer
There might be an elegant explanation of the result, but here is the bruteforce calculation.
We fix $m$ and define $E(a, b)$ as the expected number of elements to fill one table, if there are still $a$ and $b$ empty entries in the two tables.
Thus $E(0, *) = E(*, 0) = 0$ and we are looking for $E(m, m)$.
For $a, b > 0$, by considering the possibilities of the next element, we have: $$E(a, b) = 1 + \frac a {2m}E(a - 1, b) + \frac b {2m} E(a, b - 1) + (1 - \frac{a + b}{2m})E(a, b)$$ which simplifies to: $$(a + b) E(a, b) = 2m + aE(a - 1, b) + bE(a, b - 1).$$
If we define $F(a, b) = \frac 1 {2m} E(a, b)$, then we have the recurrence relation $$(a + b) F(a, b) = 1 + aF(a - 1, b) + bF(a, b - 1).$$
It is not totally obvious, but one can show by induction that $F(a, b) = H_a + H_b - H_{a + b}$ for all $a, b \geq 0$, where $H_n = \frac 1 1 + \dots + \frac 1 n$ denotes the harmonic number.
Proof: We simply need to verify that $$(a + b)(H_a + H_b - H_{a + b}) = 1 + a(H_{a - 1} + H_b - H_{a + b - 1}) + b(H_a + H_{b - 1} - H_{a + b - 1})$$ which is a routine calculation.
Therefore we have $F(m, m) = 2H_m - H_{2m}$ and hence $$E(m, m) = 2m F(m, m) = 4mH_m - 2mH_{2m}.$$