We also know $g=8$, or equivalently $$a+b+c+d+e+f=26-8=18.$$
By summing $$\color{blue}{b}+f+\color{red}{c}=6,$$ $$\color{green}{a}+e+\color{red}{c}=12,$$ $$\color{green}{a}+d+\color{blue}{b}=5$$ we're essentially double-counting $a$, $b$ and $c$ (those who answered one question) whereas we're single-counting $d$, $e$ and $f$ (those who answered two questions). This imbalance between $a+b+c$ and $d+e+f$ can be exploited: it enables us to separate $a+b+c$ out from the first equation.
The rest is just arithmetic. Let $X=a+b+c$, then the final equation implies $X+23-2X+8=26$, and we solve for $X$.
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
The numbers in the diagram indicate how many times the given formula counts that particular portion of the Venn Diagram. In the first picture, for example, an element in the intersection of A, B, and C will be counted 3 times. Each term of the inclusion-exclusion formula gradually refines these numbers until each portion of the Venn Diagram is counted exactly once.