The quotient is $q(x) = -2x^2 + 2x - \frac{5}{2}$ and the remainder is $r(x) = 5x + \frac{13}{2}$. (Sorry, I don't know a good way to typeset the calculation.) A common error was caused by forgetting to write the $0x^3$ term in $f(x)$ as a placeholder.
For a ring $R$, let $U(R)$ denote the group of units of $R$ under multiplication. For a field $F$, we have $U(F) = F^*$ because every nonzero element is a unit. Recall that for a finite field $F$, the group $F^*$ is cyclic.
One generator of the group $\mathbb{Z}_7^*$ is $3$: the successive powers of $3$ are $1$, $3$, $2$, $6$, $4$, $5$. (There is one other generator, namely $5$.)
$U(\mathbb{Z}_8)$ is not cyclic. Its elements are $1$, $3$, $5$, and $7$, and we have $1^2 = 3^2 = 5^2 = 7^2 = 1$, so none of these is a generator. (Also, this implies that $U(\mathbb{Z}_8)$ is isomorphic to the Klein four-group.)
One generator of $U(\mathbb{Z}_{10})$ is $3$: the successive powers of $3$ are $1$, $3$, $9$, $7$. (There is one other generator, namely $7$.)
By evaluating $f(x)$ at various elements of $\mathbb{Z}_5$, we find that $2$ is a zero of $\mathbb{Z}_5$. So $x+3$ (which is the same polynomial as $x-2$) divides $f(x)$ by the factor theorem. Dividing $f(x)$ by $x+3$, we get a quotient of $q(x) = 2x^2 + x + 1$. By evaluating $q(x)$ at all elements of $\mathbb{Z}_5$, we find that it does not have any zeroes. Because its degree is $2$, this means it is irreducible in $\mathbb{Z}_5[x]$. So we can factor $f(x)$ into irreducible polynomials as $f(x) = (x+3)(2x^2 + x + 1)$.
Let $f(x) \in \mathbb{Z}[x]$, say $f(x) = \sum_{i=0}^n a_ix^i$ where $a_n \ne 0$.
Assume that the rational number $p/q$ is a zero of $f$. Then $\sum_{i=0}^n a_i p^i/q^i = 0$. Multiplying both sides by $q^n$, we get $\sum_{i=0}^n a_i p^i q^{n-i} = 0$. Subtracting all terms except $a_0q^n$ from both sides, we get $a_0q^n = \sum_{i=1}^n a_ip^iq^{n-i}$, which is divisible by $p$. Because $p$ is relatively prime to $q^n$, this implies that $a_0$ is divisible by $p$. Similarly, subtracting all terms except $a_np^n$ from both sides, we get $a_np^n = \sum_{i=0}^{n-1} a_ip^iq^{n-i}$, which is divisible by $q$. Because $q$ is relatively prime to $p^n$, this implies that $a_n$ is divisible by $q$.
Assume that $f(x)$ is monic and let $p/q$ be a rational zero of $f(x)$. Without loss of generality, we may assume that $\gcd(p,q) = 1$. Then by part a, $q$ divides the leading term of $f(x)$, which is $1$. Therefore $q = \pm 1$. Therefore $p/q = \pm p$, which is an integer.
Let $m$ be a positive integer that is not a square. By definition, $\sqrt{m}$ is a root of the monic polynomial $x^2 - m$. Assume toward a contradiction that $\sqrt{m}$ is rational. Then by part b, $\sqrt{m}$ is an integer, so by squaring we see that $m$ is a square, a contradiction.
For each of the following subsets of the polynomial ring $\mathbb{Q}[x]$, say which case applies: (a) it is an ideal, (b) it is a subring but not an ideal, or (c) it is not a subring.
$\{f(x) \in \mathbb{Q}[x] : \deg f(x) \le 0\}$ is a subring but not an ideal: it contains $1$ but not $1 \cdot x$.
$\{f(x) \in \mathbb{Q}[x] : \deg f(x) \le 1\}$ is not a subring because it is not closed under multiplication: it contains $x$ but not $x \cdot x$.
$\{f(x) \in \mathbb{Q}[x] : \deg f(x) \ge 1 \text{ or } f(x) = 0\}$ is not a subring because it is not closed under addition: it contains $x$ and $1-x$, but not their sum.
$\{f(x) \in \mathbb{Q}[x] : f(x) \text{ has only integer coefficients}\}$ is a subring but not an ideal: it contains $1$ but not $1 \cdot 1/2$.
$\{f(x) \in \mathbb{Q}[x] : (x^2 + 1) \mid f(x)\}$ is an ideal.
$\{f(x) \in \mathbb{Q}[x] : f(x) \text{ is monic}\}$ is not a subring because it is not closed under addition: it contains $1$ but not $1+1$.
Let $R$ and $R'$ be rings and let $\phi : R \to R'$ be a homomorphism.
Let $N'$ be an ideal of $R'$. Then $\phi^{-1}[N']$ is an ideal of $R$:
Let $x,y \in \phi^{-1}[N']$. Then $\phi(x),\phi(y) \in N'$, so $\phi(x+y) = \phi(x) + \phi(y) \in N'$, so $x+y \in \phi^{-1}[N']$.
$\phi(0_R) = 0_{R'} \in N'$, so $0_R \in \phi^{-1}[N']$.
Let $x \in \phi^{-1}[N']$. Then $\phi(x) \in N'$, so $\phi(-x) = -\phi(x) \in N'$, so $-x \in \phi^{-1}[N']$.
Let $x \in \phi^{-1}[N']$ and $y \in R$. Then $\phi(x) \in N'$, so $\phi(xy) = \phi(x)\phi(y) \in N'$ and $\phi(yx) = \phi(y)\phi(x) \in N'$, so $xy,yx \in \phi^{-1}[N']$.
No, for example if $\phi : \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ is given by $\phi(a) = (a,a)$, then $\phi[2\mathbb{Z}]$ is not an ideal of $\mathbb{Z} \times \mathbb{Z}$: it contains $(2,2)$ but not $(2,2)\cdot (1,0)$.
(Note that the answer is "yes" under the additional assumption that the homomorphism $\phi$ is surjective. More generally, we can say that if $N$ is an ideal of $R$ then $\phi[N]$ is an ideal of $\phi[R]$. The problem only occurs when we multiply elements of $\phi[N]$ by elements of $R'$ outside $\phi[R]$.)