The question stipulates that karate is on the second floor with exactly two other teams.  Karate must be with track, according to the first rule.  The second rule effectively places either football or golf with karate.  That means the configuration must be one of two options:  karate, track, and football; or karate, track, and golf.  Baseball and swimming, therefore, must be together on another floor (just not the one with hockey).

This is a simple "grab-a-rule" question. The condition that if the seal is first then the koala is fourth eliminates one choice. The condition that if the flamingo is first then the dog is fourth eliminates two choices. And the fact that the dog must come before the cat eliminates the final wrong choice.

If there are three blue toys they must be spread out, in spots one, three and five, since no blue toys can be adjacent to each other. This leaves only two pink toys, one of which is the flamingo. Knowing this we can assume two different diagrams: one in which the flamingo is second, and one in which it is fourth. In this second diagram the flamingo would, as always, be pink, which is why this is the correct answer.

The seal cannot be first because either the flamingo is in the fourth spot where the koala should be, or if the flamingo is second and the koala can go fourth, that forces the dog and cat to both be blue. This is also the reason why the dog and cat cannot be third and fifth, respectively (they would both be blue).

If a blue cat is second, then we know that a pink dog must be first (because the dog must come before the cat, and they must be different colors). Since there must be at least two blue toys, the second blue toy is either fourth or fifth. We are left with three toys (flamingo, seal, koala) the only rule concerning them that we need to worry about is that the flamingo is pink, and therefore cannot go in whatever spot is designated for the second blue toy.

If the seal and koala are both pink, it forces the flamingo to be blue, which does not work. If the seal and the koala are both blue that means there are three blue toys total, which does not work. Same thing is the cat and flamingo are the only pink toys.

If the flamingo is first we know that there are only two blue toys, since three blue toys 2-5 would force at least two of them to be adjacent. We also know that the dog must be in the fourth spot according to the second conditional. If the dog is fourth, then we know that the cat must be fifth, since it has to come later in the line than the dog. This leaves the koala and the seal to interchangeably fill the second and third spaces.

The two blue toys bust be spread out among the second through fifth spaces. Therefore the blue toys can be second and fifth, second and fourth, or third and fifth.

This is a tricky question because of the way it is phrased (not uncommon on the LSAT).  It does NOT ask which item MUST be false, but only which one COULD BE false.  The best way to proceed is to eliminate the ones that must be true (which is really what the question is calling for--identifying what must be true).  If D is served on Monday, then A must be served on Tuesday and C must be served on Wednesday.  Why?  Because C must be served on free-wine day (Tues or Wed), since D is not an option given that it is served on Monday, and A must precede C.  We also know for sure that E is served on Friday, since Friday is reserved only for D or E, but D is taken up for Monday.  Thus, we can eliminate all the choices except for the proposition that B is served on Saturday.  Note that B could be served on Saturday, but it is not required.  Therefore, that proposition COULD BE FALSE.

Caleb doesn't like orange or yellow, so he can't be playing with those trains.  Dan is playing with the blue train.  Everett must be either next to Dan so that Everett is playing with the green train, or two spaces away, in which case Amanda must be playing with the green train because that would be the only way she could be between two boys.  That leaves the red train.

Dan is playing with the blue train.  Everett is playing with either the green or the orange train.  Caleb is playing with the red train.  Amanda is in between two boys so she can't be playing with the yellow train on the far right.  Beatrice is the only child who could be playing with the yellow train.

Dan must be playing with the blue train.  Beatrice must be playing with the yellow train.  That excludes all other answer choices but one, which also meets all of the conditions.

Dan is playing with the blue train.  In order for Amanda and Beatrice to be three spaces apart from each other, they must play with the green and the yellow train.  Amanda can't be on the end because she's in between two boys, so Amanda must be playing with the green train and Beatrice must be playing with the yellow train.  Everett can't be separated from Dan by more than one train, so he must then be playing with the orange train.  That leaves Caleb playing with the red train.