= Solution
The <Riemann-Hurwitz formula> for a nonconstant degree-$n$ morphism
$f:C\to C'$ of smooth projective connected curves in characteristic zero is
$$
2g(C)-2=n(2g(C')-2)+\deg R_f,
$$
where $R_f$ is the effective ramification divisor. If $g(C')\geq2$ and
$g(C)<g(C')$, the right side is at least $2g(C')-2>2g(C)-2$, impossible. If $g(C')=1$, then $g(C)=0$ and the left side is $-2$ while the right side is nonnegative. Thus every morphism is constant.
Choose $p\in C_d$ and project from $p$ to the pencil of lines through it. Bézout's theorem says that a general such line meets $C_d$ in $p$ plus $d-1$ further points, so the resolved projection
$$
\phi:C_d\to\mathbb P^1
$$
has degree $d-1$. Since the <genus of a smooth plane curve> is
$g=(d-1)(d-2)/2$, Riemann--Hurwitz gives
$$
\deg R_\phi
=2g-2+2(d-1)
=(d-2)(d+1).
$$
Every branch point receives at least one ramification point, hence
$$
\boxed{|B|\leq(d-2)(d+1)}.
$$
The adjunction formula gives
$$
K_{C_d}\sim(d-3)D.
$$
Since $\deg D=d$ and $\deg K_{C_d}=d(d-3)$, linear equivalence
$D\sim K_{C_d}$ would force $d=d(d-3)$, hence $d=4$. It is therefore impossible for $d\geq5$.
For a smooth plane quartic, projection from a point of the curve gives a degree-three map, so its <gonality> is at most three. Its genus is three and
$K_C\sim D$, so its plane embedding is the <canonical map>. A genus-three curve with a degree-two map to $\mathbb P^1$ would be <hyperelliptic curve>[hyperelliptic], and its canonical map would factor through that double cover and map onto a conic rather than embed the curve. Thus no degree-two map exists, and a degree-one map is excluded by the genus. The quartic's gonality is
$$
\boxed{3}.
$$
For a genus-one curve, Riemann--Roch supplies a degree-two map to $\mathbb P^1$, while degree one would make it isomorphic to $\mathbb P^1$. Its gonality is therefore
$$
\boxed{2}.
$$
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