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.
It is a matter of what is used in your course, but because the domain is the set of all people, I would leave out the $\in X$ everywhere.
For (b), it should be something like $\forall x((T(x)\land S(x))\rightarrow A(x))$.
For (d), the following will work:
$$\exists x\exists y\exists z(T(x)\land T(y)\land T(z)\land \lnot E(x,y)\land \lnot E(y,z)\land \lnot E(z,x) \land \lnot S(x)\land \lnot S(y)\land \lnot S(z)).$$
(Formally, this is not quite right, since we need many more parentheses.)
Best Answer
The first two expressions are a start, but you need to quantify $y$, where $y$ the somebody who is questioned!
$$\exists x \exists y[P(x) \land Q(x, y)]$$
$$\exists x\exists y[\lnot P(x) \land Q(x, y)]$$
Third expression:
$$\forall x[\exists yQ(x, y) \rightarrow R(x)]$$
Conclusion:
$$\exists x \exists y[(R(x) \land R(y) \land x\neq y]$$