This is almost impossible without context. I can only give an answer in general or conditional terms for many of these.
It is important to realize that checking the pKas of these compounds in relation to the pKas of other compounds in solution helps you determine where the lewis base is. Checking electronegativities to identify electrophilic/electropositive atoms also helps.
Generally, since #"NH"_3# has a pKa of #36#, and since it has a lone pair of electrons, it tends to act as a nucleophile, but only if the electrophile is electrophilic enough (such as acetone, or a weaker base than #"NH"_3#). Check your pKas.
Generally, since #"NH"_4^(+)# has a pKa of #9.4#, and since it has a positive charge, it tends to act as a lewis acid (or electrophile), donating that proton to a lewis base, but only if that lewis base is a stronger base than #"NH"_3# (the equilibrium lies on the side of the weaker acid or stronger base). Check your pKas.
#"H"^(+)# is one of the only electrophiles that is guaranteed to be an electrophile. It has no electrons, so of course, it can only accept electrons. Hence, it must be a lewis acid, or electrophile.
#"OH"^(-)# is almost always going to be a nucleophile, as it is negatively charged. Granted, sometimes it acts like a base (like with acetone), and sometimes it acts like a nucleophile (like with alkyl halides), but in general it tends to want to donate electrons.
As always, check the pKa and compare it to the molecules it reacts with. It's not going to steal a proton from #"NH"_3#, that's for sure. The pKa of water is #15.7#, so water is a stronger acid than #"NH"_3#.
You can figure out how #"H"_2"O"# can act; again, compare the pKa to the molecules it reacts with. It might be either one, but not both at the same time (unless it is interacting with itself. Can you write that equation out?).
#"BCl"_3# is the one that might be tricky. It is actually an electrophile, since boron has an empty #2p_z# orbital perpendicular to its plane that can accept electrons. You can see a similar case with #"AlCl"_3#.