The correct, formal reasoning is given by @Sachin Mittal 1 in the answer below, But I just want to point out that in the question, $h_1,h_2,$ two simple uniform hash functions are given only to confuse students.

Both $h_1,h_2$ are Simple Uniform Hash Functions, So, using any of them, for any key $k,$ each slot is equally likely. $i.e.$ **Whatever key** $k$ we take, on this key, **Whatever hash function** $(h_1\,\,or\,\,h_2)$ we apply, each bucket/slot is equally likely to be occupied.

So, for each slot, the probability that it will be occupied for any key is $1/m.$

Hence, the expected number of keys in a slot is $n/m.$

$\color{blue}{\text {NOTE 1 : }}$ The statement “suppose our hashing scheme uses $h_1$ for the odd keys and $h_2$ for the even keys” is irrelevant for uniform hash functions. Even if our hashing scheme uses $h_1$ for the $r$ keys and $h_2$ for the $n-r$ keys, we still have $1/m$ probability for any slot to be occupied for any key.

So, **Don’t think** that answer is $n/m$ because we apply $h_1$ for odd keys and $h_2$ for even keys. Assume that $n=100, m=20,$ and that these $n$ keys are such that $5$ of them are odd, and remaining all are even, then also expected number of keys in a slot will be $n/m = 100/20 = 5.$

$\color{blue}{\text {Theorem : }}$

If $H = \{ h_1,h_2,h_3,..,h_r \}$ is any collection of uniform hash functions, such that our hashing scheme uses $h_1$ for $n_1$ keys, $h_2$ for $n_2$ keys, and so on, $h_r$ for $n_r$ keys, where $n_1+n_2+..+n_r = n,$ then the expected number of keys in a hash table slot is $n/m.$

**Some Theory :**

We want to store a set of $n$ keys. A hash table is an array $T[1 .. m],$ where $m$ is another positive integer,

which we call the table size, represents number of hash table slots. Let $x$ be any key, and let $i, where \,(1<=i<=m)$, be any slot.

The definition of a “simple uniform hash function $h$” is as following :

$\color{red} {\text{Uniform hash function definition : } } Pr[h(x) = i] = 1/m$

Let me emphasize that this condition must hold for EVERT key $x$ and EVERY hash index $i.$

We say that key $x$ hashes to the slot $T[h(x)].$

If we have a **Set $H$ of such simple uniform hash functions, **then whatever hash function $h \in H$ we choose for any key $x$, it makes any slot $i$ equally likely to get occupied i.e.

$\text{for any } h \in H; Pr[h(x) = i] = 1/m, \text{for all } x \text{ and for all } i$