The book solution is missing a quantifier on $Y$. I can't see any reason for this except that it must be a typo.
All three solutions involve use of commas, which is not standard punctuation in logic, and may lead to confusion and mistakes if you use them without being very careful about which logical content you mean by them. It is much better to write exactly what you want with actual logical connectives.
Written out completely, your first formula would be
$$\tag{1} \forall X.\forall Y.(\operatorname{mother}(X) \land \operatorname{daughter}(Y,X)\to\operatorname{love}(X,Y)) $$
and I would say this is a correct solution, under the assumption that you're working with a uniary "mother" predicate rather than, say, a "female" predicate to distinguish mothers from fathers. It's a bit asymmetric to have "daughter" relation that tells about the gender of the offspring but not of the parent, so arguably a nicer representation would be
$$\forall X.\forall Y.(\operatorname{female}(X) \land \operatorname{female}(Y) \land \operatorname{offspring}(Y,X)\to\operatorname{love}(X,Y)) $$
but the choice of which predicates to use in the model is something different from how to use the chosen predicates to write formulas, and it is probably the latter that is the point of this exercise.
The formula (1) above is equivalent to
$$\tag{2} \forall X.\big(\operatorname{mother}(X) \to \forall Y.(\operatorname{daughter}(Y,X)\to\operatorname{love}(X,Y))\big) $$
which I thought was what you mean by your second formula before editing the question.
But notice that writing everything out, one of your commas has become a $\land$ and the other became a $\to$. Commas as really don't work well as do-what-I-mean logical operators. (And they won't be accepted in a classroom setting where the point of the exercise is to show that you actually know what you mean).
With a $\land$ instead of $\to$ you get
$$\tag{3} \forall X.\big(\operatorname{mother}(X) \land \forall Y.(\operatorname{daughter}(Y,X)\to\operatorname{love}(X,Y))\big) $$
which is not the same as (1) and (2). It says, instead that everybody is a mother and loves whichever daughters she has.
And your formula in the edited question
$$\tag{4} \big(\forall X.\operatorname{mother}(X) \land \forall Y.\operatorname{daughter}(Y,X)\big)\to\operatorname{love}(X,Y) $$
doesn't work at all -- it puts the quantifiers inside the parentheses, so they don't range over $\operatorname{love}(X,Y)$ at all. This is the same as saying
$$\tag{4'} \big(\forall Z.\operatorname{mother}(Z) \land \forall W.\operatorname{daughter}(W,Z)\big)\to\operatorname{love}(X,Y) $$
which says "If every two random persons are mother and daughter then $X$ loves $Y$" and depends on the context for telling us who $X$ and $Y$ are.
Edit: The question was updated; this answer refers to the original posting.
You haven't specified the meanings of Howl, Have, etc. I have made the most charitable assumptions, but if I'm wrong you may need to make some changes (for instance, if "Howl" doesn't have "at night" baked in).
- This is probably fine.
- Your two options are logically equivalent, and they're logically equivalent to how I would translate it, but my personal preference would be for something like $\forall x\left((\exists y(\ldots))\to\ldots\right)$. Even though it's equivalent, the phrasing "has any cats" makes me think of an embedded $\exists y$.
- Your answers are odd because the sentence says "have" but you don't use the $\mathrm{Have}(x,y)$ construction.
- This is close, but your FOL form is false in a world with no cats at all, even though "Peter has either a cat or a dog" could still be true if he has a dog.
- This is probably fine.
Best Answer
Let me address the first three sentences (since they form a maximal coherent collection of sentences).
We have a use for the following five predicates:
Now we can formulate the sentences as follows:
a) $\exists x\exists y: \mathsf{Student}(x) \land \mathsf{French}(y) \land \mathsf{TakeInSpring2001}(x,y)$
b) $\forall x\forall y: (\mathsf{Student}(x) \land \mathsf{French}(y) \land \mathsf{Take}(x,y)) \implies \mathsf{Pass}(x,y)$
c) $\exists x\exists y\forall z: \mathsf{Student}(x) \land \mathsf{Greek}(y) \land \mathsf{TakeInSpring2001}(x,y) \land ((\mathsf{Student}(z) \land \mathsf{TakeInSpring2001}(z,y))\implies x = z)$
The other five shouldn't be hard once one understands the above three. A different approach would be to use more specific predicates (e.g. $\mathsf{FrenchInSpring2001}(x)$ as suggested in the comments), but that would defeat the "consistent vocabulary" requirement of the question IMO.