Ten books are made into two piles. In how many ways can this be done if books as well as piles may or may not be distinguishable? Require that neither pile be empty.
Both not distinguishable: $10=1+9=8+2=7+3=6+4=5+5$. Then there are $5$ ways.
Piles distinguishable but books not: $5 \cdot 2-1 = 9$ (we subtract $1$ since $5+5=5+5$).
Books distinguishable but piles not: I say that it must be 
$$\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}+\binom{10}{5}=637$$ where, for example, $\binom{10}{3}$ represents the different number of piles of three books which can occupy the first pile. I don't sum up to $\binom{10}{10}$ because, for example, when I am selecting piles of 4 for the first pile, I'm automatically making piles of 6 in the second pile. However, the result in the book is said to be $$\dfrac{2^{10}-2}{2}=511$$ What's the problem?

Method 1:  Suppose one of the books is Shakespeare's King Lear.  The piles are distinguished by which of the books are in the same pile as King Lear.  There are $2^9$ subsets of the other books, each of which can be placed in the same pile as King Lear except the entire pile.  Hence, the number of permissible ways of distributing the books is $2^9 - 1 = 512 - 1 = 511$. 
Method 2:  Each of the ten books is placed in the first pile or second pile, which can be done in $2^{10}$ ways.  However, the two empty piles are prohibited, so there are $2^{10} - 2$ ways to distribute the books into two non-empty labeled piles.  Since the piles are not labeled, we divide by $2$ to obtain
$$\frac{2^{10} - 2}{2} = 511$$
unlabeled piles.

Books are distinguishable but piles not:  I say that it must be 
  $$\binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \binom{10}{5} = 637$$
  where, for example, $\binom{10}{3}$ represents the number of piles of three books which can occupy the first pile.  I don't sum up to $\binom{10}{10}$ because, for example, when I am selecting piles of $4$ for the first pile, I'm automatically making piles of $6$ in the second pile.

Your reasoning is correct when you select one, two, three, or four books for the first pile since the other pile has a different number of books than the first pile.  If the piles were labeled, there would be $\binom{10}{5}$ ways of selecting five books for the first pile.  However, when we select its complement, the same five books are in the second pile.  Since the piles are not labeled, selecting a particular set of five books for the first pile counts each such selection twice, once when we select a particular set of five books and once when we select its complement.  Hence, you should have 
$$\binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \frac{1}{2}\binom{10}{5} = 10 + 45 + 120 + 210 + 126 = 511$$
