= Solution
For rational $q\in(0,1)$ and positive integer $M$, define
$$
U_{q,M}=\bigcup_{m\geq M}
\left\{f\in C^\infty([0,1]):|f^{(m)}(q)|>m!m^m\right\}.
$$
Evaluation of the $m$th derivative at $q$ is continuous in the smooth-function metric, so $U_{q,M}$ is open.
It is also dense. Given $f$ and a metric tolerance $\varepsilon>0$, choose $R\geq1$ so that the contribution of all derivatives of order at least $R$ is below $\varepsilon/2$, and choose $m\geq\max\{M,R\}$. For large $K$, perturb $f$ by
$$
h_K(x)=\delta_K\cos\left(K(x-q)-\frac{m\pi}{2}\right),
\qquad
\delta_K=\frac{\varepsilon}{4R K^{R-1}}.
$$
For every $r<R$, $\|h_K^{(r)}\|_\infty\leq\varepsilon/(4R)$, so $d(f+h_K,f)<\varepsilon$. On the other hand,
$$
h_K^{(m)}(q)=\delta_KK^m
=\frac{\varepsilon}{4R}K^{m-R+1}\longrightarrow\infty.
$$
For sufficiently large $K$, $|f^{(m)}(q)+h_K^{(m)}(q)|>m!m^m$, proving density.
The space is complete by part (b). The <Baire category theorem> therefore makes
$$
G=\bigcap_{q\in\mathbb Q\cap(0,1)}
\bigcap_{M=1}^\infty U_{q,M}
$$
dense. Put $E=C^\infty([0,1])\setminus G$. Each complement $U_{q,M}^c$ is closed and nowhere dense, so $E$ is a <meagre set>, or a set of first category. Every $f\notin E$ has the required derivative growth. This is the <generic superfactorial derivative growth at rational points>.
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