Two tips: 1) It sometimes helps to rephrase the sentence into an equivalent English-sentence that looks easier to analyze. 2) Often times, you can break down the sentence to make it easier to parse. If you have trouble wrapping your head around the sentence, try phrasing it in a slightly more suggestive way. For instance:
"Every grandparent is such that either they have only daughters, or they have exactly two sons, or they have no children."
In general, "Every $\varphi$ is such that $\psi$" gets translated into the predicate calculus as $\forall x (\varphi(x) \rightarrow \psi(x))$. Your $\varphi(x)$ here is "$x$ is a grandparent", whereas your $\psi(x)$ is "$x$ either has... (etc.)". So overall, the translation should look like this:
$\forall x(x \text{ is a grandparent} \rightarrow x \text{ either has only daughters, or exactly two sons, or is childless})$
So if you can figure out how to say "$x$ is a grandparent" and "$x$ either has only daughters, or has exactly two sons, or is childless", then you'll know how to translate the sentence.
How do you say "$x$ is a grandparent"? Basically, it amounts to saying that $x$ has some child, who also has some (other) child. So this just amounts to $\exists y (C(y,x) \wedge \exists z(C(z,y)))$. This formula (which has $x$ free btw) is your $\varphi(x)$, which goes in the antecedent of the conditional of your universally quantified sentence.
How do you say "$x$ either has only daughters, or exactly two sons, or is childless"? Well, it seems to be a disjunction about $x$, so split it up into cases: if you know the whole thing is a disjunction, you can tackle each disjunct separately and then put it all together with $\vee$s at the end. So you just need to analyze "$x$ has only daughters", "$x$ has exactly two sons", and "$x$ is childless". Hopefully, things are clear enough that you can do these on your own.
If you allow function symbols, then the answer is yes. Consider the language $\Sigma$ consisting of a binary function symbol $f$ and a binary relation symbol $R$, and let $\varphi$ be the $\Sigma$-sentence
$$\mbox{$R$ is a strict linear order and for all distinct $x$ and $y$, $f(x,y)$ is $R$-between $x$ and $y$.}$$
With only relation symbols, the answer is no: any subset of a $\Sigma$-structure $\mathcal{M}$ is also a substructure of $\mathcal{M}$ if $\Sigma$ is relational; this means that any satisfiable $\Sigma$-sentence of the form $\forall x_1,...,x_n\psi(x_1,...,x_n)$ with $\psi$ quantifier-free has a finite model (take any finite substructure of any model of $\psi$). Meanwhile it's not hard to show that any satisfiable $\Sigma$-sentence of the form $\exists x_1,...,x_n\psi(x_1,...,x_n)$ with $\psi$ quantifier-free has a finite model (pick an arbitrary model, and look at a finite substructure containing the witnesses to the sentence).
Note that the argument of the last sentence also shows that no sentence of the form $\exists y_1,..., y_m\forall x_1,...,x_n\psi(y_1,..., y_n, x_1,...,x_n)$ with $\psi$ quantifier-free. So "$\forall \exists$" is really where all the necessary complexity is, if we disallow function symbols.
Best Answer
Here's a solution to a close relative of $3$, namely $$\{x+y+x^y: x,y\ge 2\}.$$
The idea is that $x^y$ counts the number of functions from a set with $y$ elements to a set with $x$ elements.
We start with the language consisting of three unary predicates $X,Y,F$ and a ternary relation $A$. Our axioms say:
$X,Y,F$ partition the domain, and $X$ and $Y$ each have at least two elements.
$A\subseteq F\times Y\times X$. Intuitively, elements of $F$ represent functions from $Y$ to $X$, and $A(f,x,y)$ means $f(x)=y$.
For all $f,y,x_1,x_2$, we have $A(f,y,x_1)\wedge A(f,y,x_2)\rightarrow x_1=x_2$; and similarly, each element of $F$ describes a total function from $Y$ to $X$.
Now a model of the theory so far amounts to a pair of sets $X,Y$ and some family of functions between them. We have to ensure that every function "appears in the $F$-part" - at least, as long as $X$ and $Y$ are finite (note that Lowenheim-Skolem implies that we can't do this for infinite $X$ and $Y$). We can do this via a cute trick:
Putting everything we have so far together we get a single sentence whose spectrum is the desired set. Now, do you see how to shift from "$x+y+x^y$" to the desired "$x^y$"? (HINT: let some elements of the structure "do double-duty.")
EDIT: note that the general strategy in the above was, "figure out a 'combinatorial story' that the set of numbers has, then tell it possibly with auxiliary sets, then if necessary 'fold in' those auxiliary sets appropriately." The same strategy, at least as I construe it, also describes Atticus' answer addressing your other question.