I think the answers you want are (in order, top to bottom) two, three, one, zero and one.
I am basing this answer on the notion that each lone pair of electrons in the valence shell of an atom of either N, O or F represents the ability of this atom to potentially form one hydrogen bond.
The doubly-bonded oxygen uses two valence electrons to form the double bond with the carbon, but has two lone pairs remaining.
The "negativity-charged oxygen atom" as the question states, has only one electron in a covalent bond, and a total of six others (including the one that causes the negative charge) in three lone pairs. Hence, three hydrogen bonds possible.
A similar situation occurs with the nitrogen atoms. The one with a lone pair can form a hydrogen bond, but the one at upper left has used all four valence orbitals to bond, and has no lone pair for a hydrogen bond.
Hope that helps!