I will use the standard symbols $\lor$ for OR and $\lnot$ for NOT. Let’s look first at the example. You have $\begin{align*} &A\lor\lnot B\tag{1}\\ &A\lor\lnot C\tag{2}\\ &C\lor D\tag{3}\\ &A\lor\lnot D\tag{4}\;. \end{align*}$
You can resolve two of these if one of them contains $X$ and the other $\lnot X$ for some $X$. Here that gives you two possibilities, $(2)$ and $(3)$, where you have $C$ in one and $\lnot C$ in the other, and $(3)$ and $(4)$, where you have $D$ in one and $\lnot D$ in the other. Resolving $(2)$ and $(3)$ gives you $A\lor D$; it contains $D$, and $(4)$ contains $\lnot D$, so you can resolve these two to get $A\lor A$, or simply $A$. This shows that $A$ must be true.
What if we’d resolved $(3)$ and $(4)$ to get $C\lor A$? We could then have resolved that with $(2)$, getting rid of the $C$’s, to get $A\lor A$ and therefore $A$, so we’d have ended up in the same place.
Now let’s look at Problem (a). You have
$\begin{align*} &A\lor B\lor C\tag{5}\\ &\lnot A\lor\lnot B\tag{6}\\ &\lnot C\lor\lnot D\tag{7}\;. \end{align*}$
The proposition letter $D$ appears only once, so there’s no hope of resolving it away. On the other hand, $(5)$ and $(6)$ look like very good candidates for resolution. Unfortunately, they’re too good in a sense: after we ‘resolve out’ $A$, we have $B\lor C\lor\lnot B$, which is a tautology: it’s always true, on account of the $B\lor\lnot B$. Resolving out $B$ also leads to a tautology. (This always happens when you have more than one complementary pair, like $A$ and $\lnot A$ or $B$ and $\lnot B$, in the two expressions.) A tautology doesn’t tell us anything about the truth of the individual propositions mentioned in it, so it’s useless to us here.
Since that didn’t get us very far, let’s resolve out $C$ in $(5)$ and $(7)$ instead, getting $A\lor B\lor\lnot D$. And that’s pretty much a dead end: we could resolve with $(6)$, but we’d just get another tautology, and no other resolutions are available. Thus, we conclude that none of $A,B,C$ and $D$ is necessarily true or false.
If in doubt, you can verify this by brute force. Is it possible for $A$ to be true? Yes: if $A$ is true, $B$ is false, and $C$ is false, $(5)-(7)$ are true no matter what $D$ is. If $A$ is false, $B$ is true, and $D$ is false, $(5)-(7)$ are true no matter what $C$ is. Thus, any of the four atomic propositions can be either true or false when $(5)-(7)$ are true.
Problem (b) is actually a bit easier; since this is homework, I’ll give you a chance to try it on your own first. You should be able to show that one of the atomic propositions $A,B,C$, and $D$ is definitely true and that one is definitely false.