Uploaded in a picture, rather than typing it all. Note that these are unmarked questions from a sample exam. Just trying to study, but have forgotten almost everything (three major exams before this one).
i) Was pretty obvious to show - $I$ is not an ideal in this case because $[3]_{4} \times [1]_{4} = [3]_{4}$ which is not in $I$ (and hence breaks pretty much the only requirement of being an ideal).
ii) Wasn't entirely sure on the terminology - does that imply that if $n$ is prime, then the element for that $n$ is zero? (e.g. For $n=13$, $\ \ a_{13}=0$, which I guess then passes as $I$ is almost equal to $R$ except for the prime gaps)
iii and iv) Again, not so sure on the terminology. Does $a\equiv b$ mod $2$ imply $a=b+2n$, for n in the integers? If so, I attempt to look at the product of some $r\ \epsilon\ R$ and an element $x$ of $I$ such that:
$r \times x = mi \times (a+bi)=aim-bm=(b+2n)mi-bm=bmi+2nmi-bm$
Then try to show that the difference between the real and imaginary coefficients is a multiple of 2:
$(bm+2nm)-bm=2nm$, which is divisible by two. Hence $I$ is an ideal of $R$. But that assumes that my definition is correct. (And the proof for iv) would be similar, but I have not yet done it).
If $I$ is an additive subgroup of a commutative ring $R$, then $I$ is called an ideal if and only if $r \cdot i \in I$ for all $r \in R$ and $i \in I$. It is then easy to see that (a) and (b) are ideals. For (c), let $r = c + di \in \mathbb{Z}[i]$ and $x = a + bi \in I$. Then $$(c+di)(a+bi) = ac - bd + (bc + ad)i.$$ We then wonder if $ac - bd = bc + ad \mod 2$. Indeed, we have $$a(c - d) = b(c + d) \mod 2,$$ since $a = b \mod 2$ by assumption and $(c-d) = (c+d) \mod 2$. Thus $$ac - ad = bc + bd \mod 2$$ and hence also $$ac - bd = bc + ad \mod 2.$$ For (d), we use the same calculation as (c), but note that $(c-d) = (c+d) \mod 4$ is not true for all $c,d \in \mathbb{Z}$. More concretely, consider $1 + i \in I$ and $i \in R$. Then $$i(1+i) = -1 + i,$$ but $-1 \neq 1 \mod 4$. Hence $I$ is not an ideal.