[step:Verify the easy direction by explicit reduction]
We show each listed discriminant has class number one by enumerating reduced forms.
Recall the [Uniqueness of Reduced Form](/theorems/1732) theorem: every equivalence class of positive definite BQFs contains exactly one reduced form $(a, b, c)$ satisfying $|b| \leq a \leq c$ (with $b \geq 0$ in the boundary cases $|b| = a$ or $a = c$). So $h(d)$ equals the number of reduced forms of discriminant $d$.
For fixed discriminant $d < 0$, a reduced form $(a, b, c)$ with $b^2 - 4ac = d$ satisfies
\begin{align*}
b^2 &\leq a^2 \leq \frac{|d| + b^2}{3}, \qquad \text{hence } a \leq \sqrt{|d|/3}.
\end{align*}
Thus there are finitely many reduced forms to check for each $d$, obtained by enumerating $a \in \{1, 2, \ldots, \lfloor \sqrt{|d|/3} \rfloor\}$ and $b \in \{-a+1, \ldots, a\}$ with $b \equiv d \pmod 2$, and verifying that $c := (b^2 - d)/(4a)$ is an integer with $c \geq a$.
Performing this enumeration for each $d$ in the two lists of the theorem, one finds exactly one reduced form in each case. We tabulate:
\begin{align*}
d = -3:\ (1, 1, 1); \qquad d = -4:\ (1, 0, 1); \qquad d = -7:\ (1, 1, 2); \qquad d = -8:\ (1, 0, 2); \\
d = -11:\ (1, 1, 3); \qquad d = -19:\ (1, 1, 5); \qquad d = -43:\ (1, 1, 11); \\
d = -67:\ (1, 1, 17); \qquad d = -163:\ (1, 1, 41).
\end{align*}
Each of these is easily verified: $b^2 - 4ac = d$ holds, and the reducedness inequalities are satisfied. Furthermore, the enumeration over $a$ with $a \leq \sqrt{|d|/3}$ produces no other valid triple. For example, at $d = -163$, $\sqrt{163/3} \approx 7.37$, so $a \in \{1, \ldots, 7\}$; one checks each value and finds only $(1, 1, 41)$ has integer $c \geq a$.
Similarly, for the non-fundamental discriminants $-d \in \{12, 16, 27, 28\}$ one has reduced forms
\begin{align*}
d = -12:\ (1, 0, 3);\ \ d = -16:\ (1, 0, 4);\ \ d = -27:\ (1, 1, 7);\ \ d = -28:\ (1, 0, 7),
\end{align*}
each unique by the same enumeration. Hence $h(d) = 1$ for all $d$ in the lists of the theorem.
[/step]