The first statement is not true.
The way you have translated these makes them questionable.
Usually they are written as:
a) $A\cap B$ is the largest subset of both $A~\&~B$.
b) $A\cup B$ is smallest set such $A\subset A\cup B~\&~B\subset A\cup B$ or $A\cup B$ is the smallest superset of $A~\&~B$.
Proof by contradiction.
Suppose that $\exists C$ such that $C\subset A~,~C\subset B~\&~C\ne A\cap B~\&~A\cap B\subset C$.
So it must be that $\exists x\in C$ such that $x\notin A\cap B$. But that means $x\in A~\&~x\in B.$ that is a contradiction.
Yes, it *is* true that A∩B is the largest set contained in A *and* contained in B (note that this is not quite what the original post says). Another proof of this:
Suppose K is a set such that: K⊆A, and K⊆B. We need to show that K⊆A∩B.
So pick any element k in K. Since K⊆A, we have that k is in A (since this is the DEFINING property of "subset", everything in K is also in A).
Likewise, k is in B, since K⊆B.
Since k is in A, and k is in B, we have that k is in A∩B = {x in U:x is in A, and x is in B} (here U can be any set that contains A and B, but it is typical to take U to be the "universe of discourse" that is: some set large enough to contain A,B and whatever "other sets" we may be considering. At the very least, we must have AUB in U).
Since we have shown that every k in K is also in A∩B, we conclude K⊆A∩B, QED.
We can construct a "similar" proof that AUB is the SMALLEST set containing A and B. For suppose we have such a set H, with A⊆H, and B⊆H.
Here, we must show that AUB⊆H.
To do this, pick any x in AUB.
Case 1: x is in A. Since x is in A, x is in H, because A⊆H.
Case 2: x is in B. The same reasoning shows x is in H.
Since x is in A, or x is in B (or both), in all cases, x lies in H, so AUB⊆H, QED.
By the way, how you prove things depends very strongly on how the words in the statement are defined. I have seen texts in which is defined as "the largest set contained in both A and B" and is defined as "the smallest set that contains both A and B".