(I figured that the statement of my second question is true. I am sharing a proof sketch for future reference.)
Let $\mu = \mathbb{E}\left[{X}\right]$,
and define $Y = X- \mu$.
It suffices to show that
$$
\mathbb{E}
\left[
\exp\left(\lambda \cdot Y\right)
\right]
\le
\exp\left(\frac{1}{2} \cdot \sigma^{2} \cdot \lambda^{2}\right),
\quad \lambda \in \mathbb{R},
$$
for $\sigma^{2} = O(B\sqrt{\log{C}})$.
(In the sequel, I assume that $\log{C} \ge 1$.)
First, we derive an upper bound on the absolute moments of $Y$:
\begin{align}
\mathbb{E}
\left[
\left\lvert{Y}\right\rvert^{p}
\right]
=
\mathbb{E}
\left[
|{X} - \mu |^{p}
\right]
\le
\mathbb{E}\left[ \max\left\lbrace {X}, \mu\right\rbrace^{p} \right]
\le
\mathbb{E}\left[ {X}^{p} \right] + \mu^{p},
\end{align}
where we have taken into account that $X\ge 0$ and in turn $\mu \ge 0$.
Now, $X^{p}$ is a nonnegative random variable, and in turn
\begin{align}
\mathbb{E}\left[ X^{p} \right]
&=
\int_{0}^{\infty} \mathrm{P}\left( {X}^{p} \ge u\right) du
=
\int_{0}^{\infty} \mathrm{P}\left( {X}^{p} \ge t^{p}\right) \cdot {p} \cdot t^{p-1} dt
\nonumber\\
&=
\int_{0}^{\infty} \mathrm{P}\left( {X} \ge t\right) \cdot {p} \cdot {t^{p-1}} dt
\nonumber\\
&=
\underbrace{
\int_{0}^{t_{0}} \mathrm{P}\left( {X} \ge t\right) \cdot {p} \cdot {t^{p-1}} dt
}_{I_{1}}
+
\underbrace{
\int_{t_{0}}^{\infty} \mathrm{P}\left( {X} \ge t\right) \cdot {p} \cdot {t^{p-1}} dt,
}_{I_{2}}
\label{sug-gaussian:X-pth-moment-start-ub}
\end{align}
for any $t_{0} \ge 0$.
For $t_{0} = ({B}\sqrt{\log{C}})^{1/2}$, we have
\begin{align}
I_{1}
=
\int_{0}^{t_{0}} \mathrm{P}\left( {X} \ge t\right) \cdot {p} \cdot {t^{p-1}} dt
\le
\int_{0}^{t_{0}} {p} \cdot {t^{p-1}} dt
=
t_{0}^{p}
=
\left({B}\sqrt{\log{C}}\right)^{p/2}.
\label{sug-gaussian:I1-ub}
\end{align}
For the second part,
let $f(t) = {C} \cdot e^{-t^{2}/B}$ and $g(t) = e^{-t^{2}/\left({B}\sqrt{\log{C}}\right)}$.
One can verify that $f(t) \le g(t)$ for $t \ge ({B}\sqrt{\log{C}})^{1/2}$ (assuming that $\log{C} \ge 1$).
Then,
\begin{align}
I_{2}
&=
\int_{t_{0}}^{\infty}
\mathrm{P}\left( {X} \ge t\right) \cdot {p} \cdot {t^{p-1}} dt
\le
\int_{t_{0}}^{\infty}
f(t) \cdot {p} \cdot {t^{p-1}} dt
%\nonumber\\&
\le
\int_{t_{0}}^{\infty}
g(t) \cdot {p} \cdot {t^{p-1}} dt,
\label{sug-gaussian:I2-ub}
\end{align}
It follows that for the particular choice of $t_{0}$,
\begin{align}
I_{2}
&\le
\int_{t_{0}}^{\infty}
g(t) \cdot {p} \cdot {t^{p-1}} dt
\le
\int_{0}^{\infty}
g(t) \cdot {p} \cdot {t^{p-1}} dt
\nonumber\\
&\le
\int_{0}^{\infty}
e^{-t^{2}/(B\sqrt{\log{C}})} \cdot {p} \cdot {t^{p-1}} dt
=
(B \sqrt{\log{C}})^{p/2}
\cdot \frac{p}{2}
\cdot \Gamma\left( \frac{p}{2}\right)
\end{align}
Combining the bounds on the two parts, we have
\begin{align}
\mathbb{E}\left[ X^{p} \right]
\le
\left({B}\sqrt{\log{C}}\right)^{p/2}
\cdot
\left(
1 + \frac{p}{2} \cdot \Gamma\left( \frac{p}{2}\right)
\right).
\end{align}
Specifically for $p=1$, we find
\begin{align}
\mu
=
\mathbb{E}\left[ X\right]
\le
2 \cdot \left({B}\sqrt{\log{C}}\right)^{1/2}.
\label{sub-gaussian:ub-on-mean}
\end{align}
Further, for any $p \ge 2$,
\begin{align}
\mathbb{E}\left[ X^{p} \right]
\le
\left({B}\sqrt{\log{C}}\right)^{p/2}
\cdot 2^{p}
\cdot \left( \frac{p}{2}\right)^{p/2}.
\end{align}
Finally, combining the above we find that for any $p \ge 2$,
$$
\mathbb{E}
\left[
\left\lvert{Y}\right\rvert^{p}
\right]
\le
c^{\prime}
\left({B}\sqrt{\log{C}}\right)^{p/2}
\cdot 2^{p}
\cdot \left( \frac{p}{2}\right)^{p/2},
$$
for some positive constant $c^{\prime}$.
Having the above upper bound on
$\mathbb{E}\left[ |Y|^{p}\right]$,
we can now show (see Lemma 5.5 in https://arxiv.org/pdf/1011.3027v7.pdf that
\begin{align}
\mathbb{E}
\left[
\exp\left({\lambda}{Y}\right)
\right]
&=
\exp(O\left((B\sqrt{\log{C}})^{2}\lambda^{2}\right))
\end{align}
which is the desired result.
