Diagram the first premise: $A \subset B$. This means that there cannot be anything inside of $A$ that is outside of $B$. In a Venn diagram, you do this by shading the area inside of $A$ and outside of $B$ ... the shading means that that area is empty:
OK, now we add to this the second premise, which is that $B \cap C = \emptyset$. So this time, the intersection of $B$ and $C$ needs to be empty, i.e. shade that very intersection. We add this to the diagram:
This diagram represents the truth of the premises. The question is now: does this force the conclusion to be true? Well, the conclusion states that $A \cap C = \emptyset$, and if you look at the diagram, we find that indeed the intersection of $A$ and $C$ is shaded, i.e. is empty. So yes, the conclusion has to be true given the diagram, i.e. given the truth of the premises. So, this is a valid argument.
For the second problem, again start with the first premise: $C \subset A \cup B$. This means that there cannot be anything in $C$ that is outside of both $A$ and $B$, and so we shade that area:
Now for premise 2: $A \cap B \cap C = \emptyset$. So, we shade the intersection of $A$, $B$, and $C$:
Now, we ask the question: does this diagram, representing the truth of the premises, force the conclusion to be true? The conclusion says that $A \cap C = \emptyset$. So, is the intersection of $A$ and $C$ empty? Well, it could be ... but there can also be something $X$ that is in the intersection of $A$ and $C$ but outside $B$:
As such, we can quickly generate a counterexample: we need to have something that is shared by $A$ and $C$, but not $B$. ... while it should still be true that there is nothing in $C$ outside of $A$ and $B$, and nothing in the intersection of all three. OK, easy:
$A = C = \{ bananas \}$
$B = \emptyset $
Best Answer
A Venn diagram with three sets $A,B,C$ divides the universe into 8 distinct regions, as you can see in the picture below.
For example, the $A$ circle contains 4 regions: the upper region with an arrow pointing to it represents $A=true, B=false,\text{ and }C=false$, since that region is contained in $A$ but not in $B\text{ or }C$ For your function, the truth table already identifies the truth values of the points in each of the 8 regions. For each of these, place a marker in those regions where your truth table says the function evaluates to $true$, as I did in the other region pointed to by an arrow. Your function is $true$ in four cases; I've put a dot in each. (Your conventions for marking regions might differ from mine: you might want to shade them rather than placing a dot.)