This looks like homework, so I'll just guide you through the las problem and see if you can do the rest on your own.
d) j: {words} → {letters}, j(x) = initial letter of x
Domain: words Co-Domain: letters Range: letters
Is this onto? Because there is at least one word that starts with each letter, you can get all possible values in the Co-Domain, so it is onto.
Is this one-to-one? Well, no. For example, apple and alligator will give the same return value even though they are different words.