As the other colleague mentioned, for these proofs you should directly use the definitions of intersection and union that you got in class, and try and land on what you need from there. Let's decompose it to the bone, for a first time.
The general pattern for showing that $ A = B$, with $A$ and $B$ sets, is to show that $A \subset B$ and $B \subset A$. Try and draw sets as circles on a piece of paper : you will see that the only way to include a set in another, and vice-versa, is to have them be equal.
Generally still, to prove that $ X \subset Y$, with $X$ and $Y$ sets, one way is to take any $x$ in $X$ and show that $x$ is also in $Y$ (call this concept number (1)). Since $x$ is any element of $X$ then this applies to all elements of $X$ and so all of $X$ is contained in Y.
So, let's do exactly that for problem a) : let $x$ be any element of $A \cap (B \cup C)$. Now we want to show that $x \in (A \cap B) \cup (A \cap C) $ right ? (This is concept (1)) Well, what does this mean ?
It means, by definition of $\cup$, that $x \in (A \cap B)$ OR $x \in (A \cap C)$ (let's call this equation (2)). We'll see if this equation (2) is true.
Now what do we know about $x$ ? Well, since $x \in A \cap (B \cup C)$, it follows that, by definition of $\cap$, $x \in A$ AND $x \in (B \cup C)$. That first part, we'll call number (3). That last part can be rewritten as $x \in B$ OR $x \in C$ - it's gotta be one of the two.
Suppose for example that $x \in B$. Since we also know (from number (3)) that $x \in A$, it follows by definition of $\cap$ that $x \in (A \cap B)$. In this case then, we have equation (2) fulfilled.
Now suppose that $x \in B$. Again using number (3), we know that $x \in A$, and by definition of $\cap$, $x \in (A \cap C)$ and we have once again fulfilled equation (2).
So this means that in all possible cases for $x$, we have equation (2) fulfilled and true - $x$ is indeed either in $(A \cap B)$ or $A \cap C)$.
Using concept (1), we can now say that since this applies to any $x$ in $(A\cap (B\cup C))$, then $(A\cap(B\cup C)) \subset (A\cap B)\cup (A\cap C)$.
Care to try the other direction, i.e. $(A\cap B)\cup (A\cap C) \subset (A\cap(B\cup C))$ ? Then using concept (0) you will have shown that your sets are equal.
PS : I use $ \subset$ for $ \subseteq $ indifferently. The concept of $A \subsetneq B$, i.e. "$ A\subset B$ AND $A \neq B$" is not often used anyway, so confusions rarely happen.