Fairly simple question
In a country with $x$ people I choose two subsets sets, $A, B$ with uniform distribution. The sets I choose are of fixed size $|A| = a, |B| = b$ which i knew before choosing the sets.
What is the expected value of $|A \cap B|$?
For each person $h$ of that country, define the random variable $X_h$ as $1$ if $h$ is in $A\cap B$, and $0$ if not.
What is the expected value of $X_h$ for a given $h$? It's clearly
$$E[X_h]=1\times P(h \in A \cap B) + 0 \times (1-P(h \in A \cap B)) = P(h \in A \cap B).$$
Since $A$ and $B$ are chosen indepently from each other (that isn't explicitely stated, but seems a natural assumption), we have
$$P(h \in A \cap B) = P(h \in A)P(h \in B).$$
Because the sets are chosen uniformly from the subsets with cardinality $a$ and $b$, resp., this means
$$P(h \in A \cap B) = \frac{a}x\frac{b}x = \frac{ab}{x^2}.$$
The key observation now is that the expected value you seek equals the expected value of the sum of all those $X_h$ (sum taken over the set of all persons in this country, which I denote with $H$), and that that latter sum can be split into the sum of expected values (see linearity of expectation)
$$E[|A\cap B|]=E[\sum_{h \in H}X_h] = \sum_{h \in H}E[X_h] $$
In the latter sum each summand has the same value which we calculated above, so we finally get
$$E[|A\cap B|]=x\times \frac{ab}{x^2} = \frac{ab}{x}.$$
The expected value is just the probability that someone is counted in both sets. The probability of being in set $A$ is $a/x$ and in $B$ is $b/x$, so the probability of being in both sets is $ab/x^2$.
You can verify this by considering simple cases like $a=b=x$ where we get $1$, or $x=2$ and $a=b=1$ where we get $1/4$.
