The answer to the title question is no and the answer to the body question is yes.
First, Stickelberger's theorem asserts that the discriminant $\Delta_K$ of a number field $K$ is congruent to $0, 1 \bmod 4$. So that already rules out half of the possibilities.
Second, Minkowski's bound implies that if $K$ has degree $n$ then
$$\sqrt{|\Delta_K|} \ge \left( \frac{\pi}{4} \right)^{n/2} \frac{n^n}{n!}.$$
The RHS is always strictly bigger than $1$, so the discriminant can also never be equal to $1$; that is, $\mathbb{Q}$ has no unramified extensions. Moreover it follows that number fields of a given discriminant cannot have arbitrarily high degree, and in fact there are only finitely many number fields of a given discriminant, so for particular small discriminants one can rule them out via casework.
To start, let's record the value of the Minkowski bound for some small values of $n$.
- For $n = 2$ it is about $1.57$, so a number field of degree at least $2$ has discriminant of absolute value at least $3$.
- For $n = 3$ it is about $3.13$, so a number field of degree at least $3$ has discriminant of absolute value at least $10$.
- For $n = 4$ it is about $6.58$, so a number field of degree at least $4$ has discriminant of absolute value at least $44$.
Let's also recall that for a squarefree integer $d$ the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$ if $d \equiv 1 \bmod 4$ and $4d$ otherwise. Now let's go through the smallest few discriminants in order, skipping the ones that are impossible by Stickelberger to see what Minkowski has to say about them.
- $1$: impossible by Minkowski.
- $-3$: realized uniquely by $\mathbb{Q}(\sqrt{-3})$.
- $4$: impossible by Minkowski (can only be realized by a quadratic field and isn't).
- $-4$: realized uniquely by $\mathbb{Q}(i)$.
- $5$: realized uniquely by $\mathbb{Q}(\sqrt{5})$.
- $-7$: realized uniquely by $\mathbb{Q}(\sqrt{-7})$.
- $8$: realized uniquely by $\mathbb{Q}(\sqrt{2})$.
- $-8$: realized uniquely by $\mathbb{Q}(\sqrt{-2})$.
- $9$: impossible by Minkowski (can only be realized by a quadratic field and isn't).
- $-11$: realized by $\mathbb{Q}(\sqrt{-11})$.
- $12$: realized by $\mathbb{Q}(\sqrt{3})$.
$-12$ is the first discriminant whose status I can't determine from just Stickelberger and Minkowski. If it occurs as a discriminant it must be the discriminant of a cubic field. I think it's known that in fact the smallest possible discriminant (in absolute value) of a cubic field is $-23$, realized by $\mathbb{Q}(x)/(x^3 - x - 1)$, but I don't know how to prove this.
In any case, here's a discriminant I can rule out using an additional technique: $25$ is congruent to $1 \bmod 4$ and, by the Minkowski bound, could only be the discriminant of a cubic field. However, I claim it isn't the discriminant of a cubic field, and so can't be a discriminant at all. The reason is that it's a square, which means that the corresponding cubic field is Galois with Galois group $A_3 \cong C_3$. By the Kronecker-Weber theorem it must therefore be a subfield of the cyclotomic integers $\mathbb{Q}(\zeta_n)$, and it's known that we should in fact be able to take $n = 5$. However, $\mathbb{Q}(\zeta_5)$ has no cubic subfields since it has degree $4$.
This following example was found via a computer search for simple integral basis and small discriminant, so that the discriminant is easy to compute and the Minkowski bound is small.
I am not sure if this suffices as a good example though: The only method I know of computing class group is the basic one via Minkowski's bound and this example still involved quite a bit computation since the bound is fairly large at $<38$. I wasn't able to find a cubic extension, $\mathbb Z_2\times \mathbb Z_2$ class group with small bounds.
The stats was also checked on Sagemathcell to be sure.
Let $f(x) = x^3 + 11x+21\in\mathbb Z[x]$.
1. $f(x)$ is irreducible over $\mathbb Z$.
2. Let $\alpha \in \mathbb C$ be a root of $f(x)$ and consider the number field $K=\mathbb Q(\alpha)$.
We can show that $\{1,\alpha,\alpha^2\}$ is an integral basis and $K$ has prime discriminant $-17231$.
3. Hence Minkowski bound $$M_K=\frac{8}{9\pi} \sqrt{17231} < 38,$$ and we can find the list of non-principal ideals for each prime $\leq 37$.
4. Finally we can show that the class group $H(K)$ of $K$ is $H(K)\cong \mathbb Z_2\times \mathbb Z_2$, with generators
$$<3,\alpha>,<3,\alpha-1>$$
Edit 1: We check that $<3,\alpha>,<3,\alpha-1>$ has order 2. Let
$$
\begin{align}
I &:= <3,\alpha>^2 = <9,3\alpha,\alpha^2>\\
J &:= <3,\alpha-1>^2 = <9,3\alpha-3,\alpha^2-2\alpha+1>
\end{align}
$$
We claim that
$$
\begin{align}
<9,3\alpha,\alpha^2> &= <2\alpha^2 - 3\alpha + 27>\\
<9,3\alpha-3,\alpha^2-\alpha+1> &= <\alpha+2>
\end{align}
$$
Clearly, we have
$$
2\alpha^2 - 3\alpha + 27 \in <9,3\alpha,\alpha^2> \implies <2\alpha^2-3\alpha+27> \subseteq <9,3\alpha,\alpha^2>
$$
We obtain
$$
\begin{align}
-(\alpha^2+3\alpha+2)(2\alpha^2 - 3\alpha + 27) &= 9\\
(\alpha^2 + 6\alpha + 7)(2\alpha^2 - 3\alpha + 27) &= \alpha^2
\end{align}
$$
Therefore $9,\alpha^2\in <2\alpha^2 - 3\alpha+27>$, which in turn gives $3\alpha\in <2\alpha^2 - 3\alpha + 27>$. Hence
$$
\begin{align}
<9, 3\alpha,\alpha^2> &\subseteq <2\alpha^2 - 3\alpha + 27>\\
\implies <9, 3\alpha,\alpha^2> &= <2\alpha^2 - 3\alpha + 27>
\end{align}
$$
This shows the first equivalence. On the other hand,
$$
\begin{align}
(\alpha^2-2\alpha+15)(\alpha+2) &= 9\\
3(\alpha+2)-9 &= 3\alpha-3\\
(\alpha+2)^2-2(3\alpha-3)-(9) &= \alpha^2-2\alpha+1
\end{align}
$$
Therefore $9,3\alpha-3,\alpha^2-2\alpha+1\in <\alpha+2>$. For the reverse containment,
$$
\begin{align}
(\alpha + 2)(\alpha^2 - 2 \alpha + 1) + 5 (3 \alpha - 3) + 4 (9) &= \alpha+2
\end{align}
$$
shows that $\alpha+2 \in <9,3\alpha-3,\alpha^2 - 2\alpha+1>$. Therefore
$$
<9,3\alpha-3,\alpha^2-2\alpha+1> = <\alpha+2>
$$
Another example might be $f(x) = x^3+8x+60$ with integral basis $\{1,\alpha,\alpha^2/2\}$.
Best Answer
Here's how we go about doing (b). First, a lemma:
Lemma 1: For $\alpha \in K$, we have that $\text{Tr}_{K/\Bbb{Q}(\alpha)}(\alpha)$ and $N_{K/\Bbb{Q}(\alpha)}(\alpha)$ being algebraic integers iff $\alpha$ is integral over $\Bbb{Z}$.
Proof: Suppose that $\alpha \in \mathcal{O}_K$. Then $\alpha$ is certainly integral over $\mathcal{O}_{\Bbb{Q}(\sqrt{m})}$ and thus its minimal $m_\alpha(t)$ polynomial has coefficients in $\mathcal{O}_{\Bbb{Q}(\sqrt{m})}$. Since $\text{Tr}_{K/\Bbb{Q}(\sqrt{m})}(\alpha) = (-1) \times \text{(coefficient of $t$})$ and $N_{K/\Bbb{Q}(\sqrt{m})}(\alpha)$ is the constant term in $m_\alpha(t)$, these are both in $\mathcal{O}_{\Bbb{Q}(\sqrt{m})}$ and thus are algebraic integers. Conversely if $T = \text{Tr}_{K/\Bbb{Q}(\sqrt{m})}(\alpha)$ and $N = N_{K/\Bbb{Q}(\sqrt{m})}(\alpha)$ are algebraic integers, $\Bbb{Z}[N,T]$ is a finitely -generated $\Bbb{Z}$ - module. Also, $\alpha$ is integral over $\mathcal{O}_{\Bbb{Q}(\sqrt{m})}$ and so $\Bbb{Z}[N,T][\alpha]$ is a finitely - generated $\Bbb{Z}[N,T]$ - module. Then $\Bbb{Z}[N,T][\alpha]$ is a finitely - generated $\Bbb{Z}$ - module and so by Proposition 5.1(c) of Atiyah - Macdonald, $\alpha$ is integral over $\Bbb{Z}$.
Proposition 1: Suppose $m \equiv 1\mod{4}$, $n \equiv k \equiv 2$ or $3 \mod{4}$. Then an integral basis for $\mathcal{O}_K$ is
$$\left\{ 1, \frac{1 + \sqrt{m}}{2},\sqrt{n},\frac{\sqrt{n} + \sqrt{k}}{2} \right\}$$
where $k = mn/(m,n)^2$.
Proof: Write down the minimal polynomial for $\alpha$ over $\Bbb{Q}(m)$. Then by the lemma we see that if $\alpha \in \mathcal{O}_K$ is of the form
$$\alpha = \frac{a + b\sqrt{m} + c \sqrt{n} + d\sqrt{k}}{2}$$
with $a,b,c,d\in\Bbb{Z}$ and $a \equiv b\mod{2}$, $c\equiv d\mod{2}$. We can rewrite $\alpha$ as
$$\begin{eqnarray*} \alpha &=& \frac{a +b\sqrt{m} - b + b + c\sqrt{n} - d\sqrt{n} +d\sqrt{n} + d\sqrt{k} }{2}\\ &=& \frac{ a-b}{2} + b\left(\frac{1 + \sqrt{m}}{2}\right) +\frac{c-d}{2}\left(\sqrt{n}\right) + d\left(\frac{\sqrt{n} + \sqrt{k}}{2}\right)\end{eqnarray*}$$
Since $\frac{a-b}{2}$ and $\frac{c-d}{2}$ are arbitrary integers this concludes the proof of the proposition. Now that you have an integral basis, you only need to calculate the determinant of a $4 \times 4$ matrix to get the discriminant of $\mathcal{O}_K$.
Note to user: Since this is a homework problem I have left out some details. Among the details you need to fill in are:
How did I get that $\alpha$ must be of the form prescribed above?
How does my calculation in the proposition show that what I claimed is an integral basis for $\mathcal{O}_K$?
You are urged to fill them in and follow my method to do (a).