1a) I understand Q to be the set of rational numbers (a/b where a and b are integers) and A, in this case, to be this set excluding the number 1, that is, a/b≠1 and a≠b. (b≠0 is another condition.)
(i) The operation is commutative because a and b can be interchanged:
a:b=b:a=a+b-ab=b+a-ba. This is because addition and multiplication are themselves commutative.
(iii) A={a/b | a,b∈Z, a≠b}. So A={..., -¼, -⅓, -½, ½, ⅓, ¼, ... 2, 3, 4, ..., -2, -3, -4, ..., ⅔, ⅖, ..., 3/2, ¾, 5/2, ...} for example. A is a group. Note that, for example, 2=2/1=4/2=6/3=..., ½=2/4=3/6=... so fractions have multiple representation a/b=c/d, but a,b,c,d∈Z for some a, b, c, d.
1b)
(i) Homomorphism requires f(a*b)=f(a)*f(b) for a function f. Given φ:G→G, φ(x)=e ∀ x∈G, to show φ(x*y)=φ(x)*φ(y). φ(x*y)=e; φ(x)=e, φ(y)=e, and e*e=e by definition of identity. So φ(x*y)=φ(x)*φ(y), satisfying homomorphism.
Isomorphism is a special case of homomorphism in that it's also bijective (unique mapping).
More to follow in due course...