Probability of a Full House for five-card hand.

You are correct that your initial logic under-counts, but the correction of $5!$ is too high. In fact the correct factor to adjust by is $10.$

Before incorrectly multiplying by $5!$, you correctly compute the probability of getting the pattern (xxx)(yy) (hopefully it's clear what I mean by that). Of course this is not the only pattern a full house can come in. It can also be like (yy)(xxx), or (y)(xxx)(y), etc. The way to get the answer would be to compute the probability for all these patterns that constitute a full house, and since they're the mutually exclusive ways to get a full house, just add them up to get the probability of a full house.

Of course the probability is the same for all the patterns (though you may want to talk yourself through a weird one like (x)(y)(x)(y)(x) to make sure it is obvious why it must be the same as for your choice.) So it's just a matter of counting all the patterns. Well, there are five spots and you need to choose three to be x so that's ${5\choose 3} = 10.$

In multiplying by $5!$ you forgot that you had already taken into account the permutations of the cards that only interchange the x's and y's amongst themselves. After all, $(1)(3/51)(2/50)$ is the correct probability of drawing three of a kind in a three card hand... you wouldn't multiply that by $3!.$

Because the second card doesn't need to be another card in the triplet. It can be any other card in the deck, which starts the construction of the pair. Likewise, at every step, it doesn't need to be part of the pair or needs to be part of the triplet, it needs be part of {the pair or the triplet}.