This is pretty confused.
First: it's not enough to just give a set and ask about "ring structure" or "group structure"; you need to say what the operations are. Presumably, you want the operations to be inherited form $\mathbb{C}$ (where the values of $g(a)$ lie).
Second: the problem is incorrect as stated because you have placed no condition son $f$. Take $f(x) =x-\pi$, a polynomial that is irreducible in $\mathbb{C}[x]$ (you never said what kind of polynomial it was, and since you open by considering complex numbers, the obvious interpretation would be that $f$ is a polynomial with complex coefficients). Then $a=\pi$ satisfies the desired condition, but the set $\{g(a)\mid g(x)\in\mathbb{Z}[x]\}$ is isomorphic to $\mathbb{Z}[x]$, and its additive group is free abelian of countably infinite rank, and not isomorphic $\mathbb{Z}^1$, as the statement requires.
So presumably, you want $f(x)$ to be a polynomial with integer or rational coefficients. (Which?) But even more is required, as Pete's comment shows.
So my first hint (nay, strong suggestion), which is very applicable to all math problems is: make sure you write down all the hypothesis!
Now: part (1) has two parts: you want to show that the set is a ring (under the addition and product of complex numbers); and you want to show that as an abelian group it is free of rank $n=\deg f$.
For the first part of part 1, you need to show that the set is nonempty, and that if $g(a)$ and $h(a)$ are in the set, then so is $g(a)-h(a)$; i.e., that you can find a polynomial $k(x)\in\mathbb{Z}[x]$ such that $k(a)$ evaluates to the same thing as $g(a)-h(a)$. This shows it's a subgroup of $\mathbb{C}$. Then you need to show that it is closed under multiplication: if $g(a),h(a)$ are in the set, then so is $g(a)h(a)$; that is, there is a polynomial $k(x)\in\mathbb{Z}[x]$ such that $k$, evaluated at $a$, equals the product of the values of $g$ and $h$ when evaluated at $a$. This part should be easy.
For the second part, try showing that the group is freely generated by $1$, $a$, $a^2,\ldots,a^{n-1}$; you'll need to figure out exactly what the correct conditions on $f$ are supposed to be, otherwise this is going to be very hard (if not false; see above).
For part 2, suppose that you have a nonzero element in the ideal, and think about what happens when you multiply it by $1$, $a$, $a^2$, $a^3,\ldots,a^{n-1}$.