For the first question, you have on the one hand that $(A+B)+(C+D)=24$ and on the other hand that $A+B>C+D$. What are the possibilities for $A+B$ and $C+D$? Clearly you must have $A+B>12$ and $C+D<12$. If two single-digit natural numbers add up to $11$ or less, what are the possibilities? (Note that the answer will depend on whether $0$ counts as a natural number.)
The second question is harder. Because it’s multiple choice, however, you can solve it by elimination. First, if you set $x=y=z=\frac13$, you find that $xy+yz+zx=\frac13$; assuming that one of the answers actually does give the set of possible values, this rules out choices (A) and (B), since they both exclude $\frac13$ from the set of possible values. Next, note that
Now $x^2+y^2+z^2$ is certainly positive, so $1-(x^2+y^2+z^2)<1$, and therefore
Answer (D) therefore includes too much to be the exact set of possible values: it includes $\frac12$, which we now know isn’t possible. Thus, only (C) can be the exact set of possible values of $xy+yz+zx$.
In fact this can be proved by showing that $x^2+y^2+z^2\ge\frac13$ when $x+y+z=1$ and $x,y$, and $z$ are positive, but I don’t immediately see a way to do that without using a bit more than what I consider precalculus algebra.