For the first part , $f:A->B$ is a function from set A to set B , set A has men and set B have women and $f$ is a function that maps a man in set A to his wife in set B. [Considering each man can have atmost one wife , otherwise it wouldn't be a function].

Let $S \ and \ T$ be two subsets of $A$ , where $S$ is the set of software-engineers and $T$ be a set of Indians .

$f(S)$ gives the set of the wives of software-engineers and $f(T)$ gives the set of wives of Indians.

$f(S) \cup f(T)$ gives the set of wives of both software-engineers and Indians.

$f(S \cup T)$ also gives the set of wives of both software-engineers and Indians .

But can we say that they are same ? How can they be different? Only if $f(S \cup T)$ has less elements than $f(S) \cup f(T)$ or more than $f(S) \cup f(T)$ , only then we can say that the results are different .

Let's try taking some small sets and try to visualize , let T = {"Ramesh" , "Suresh" , "Quentin Tarantino"} , Let S={"Severus Snape" , "Gandalf" , "Suresh"} .

$f(Ramesh) = Seeta$

$f(Suresh) = Geeta$

$f(Quentin Tarantino) = Uma \ Thurman$

$f(Severus Snape) = Lilly \ Potter$

$f(Gandalf)=\phi$

$ f(S \cup T) = f(\{"Ramesh","Suresh","Quentin Tarantino"\} \cup \{"Severus \ Snape","Gandalf","Suresh"\})=$

$f(\{"Ramesh","Suresh","Quentin \ Tarantino","Severus \ Snape" , "Gandalf"\}) = \{"Seeta","Geeta","Uma \ Thurman","Lilly \ Potter"\}$

Now ,

$f(T) = \{"Seeta","Geeta","Uma \ Thurman"\}$

$f(S) = \{"Lilly \ Potter",\phi,"Geeta"\}$

$f(S) \cup f(T) = \{"Seeta","Geeta","Uma \ Thurman","Lilly \ Potter"\}$

Thus we can see that both are same .

Now for the second one , we can take $f:i->i^{2} ,\ where \ i \in I$.

We can easily show that , $f(S \cap T) \subset= f(S) \cap f(T)$ .

If we carefully take S and T then it can be proven ,

Let $S=\{2,3\}$ and $T=\{2\}$ , here $f(S \cap T) = f(2) = 4$ which is same as $f(\{2,3\}) = \{4,9\} \cap f(2)=\{4\} = \{4\}$.

How to show for proper inclusion?

Let S={-2} and T={2}.

$f(S \cap T) = f(\{-2\} \cap \{2\}) = f(\phi) = \phi $ which is not same as , $f(-2) = 4 \cap f(2) = 4 $ which is {4} .

and we know that $\phi \subset {4}$ .

Thus , proper inclusion can be proven like this.