= Solution
Use the combinations $a_\pm=a\pm b$ at each <momentum>. Part <2/a> gives $[Q,a_+]=0$, $[Q,a_-]=-2a_-$, while their <creation operators> obey $[Q,a_+^\dagger]=0$, $[Q,a_-^\dagger]=2a_-^\dagger$. Applying <2/b> gives
$$
\begin{aligned}
C_\lambda a C_\lambda^\dagger
&=\frac{1+e^{-2i\lambda}}2a+\frac{1-e^{-2i\lambda}}2b,\\
C_\lambda b C_\lambda^\dagger
&=\frac{1-e^{-2i\lambda}}2a+\frac{1+e^{-2i\lambda}}2b.
\end{aligned}
$$
The <creation operators> transform by the <Hermitian adjoint> formulas with $e^{+2i\lambda}$. At $\lambda=\pi/2$, both exponentials are $-1$, so
$$
\boxed{Ca_{\mathbf p}C^\dagger=b_{\mathbf p},\quad
Cb_{\mathbf p}C^\dagger=a_{\mathbf p},\quad
Ca_{\mathbf p}^\dagger C^\dagger=b_{\mathbf p}^\dagger,\quad
Cb_{\mathbf p}^\dagger C^\dagger=a_{\mathbf p}^\dagger,\qquad C=C_{\pi/2}.}
$$
Substituting into the fixed-time <mode expansion of a free field> of the <complex scalar field> gives
$$
C\phi(\mathbf x)C^\dagger
=\int_{\mathbf p}\frac{a_{-\mathbf p}^\dagger+b_{\mathbf p}}{\sqrt{2E_{\mathbf p}}}
e^{i\mathbf p\cdot\mathbf x}
=\phi^\dagger(\mathbf x).
$$
Thus $C$ exchanges particles and <antiparticles>, as required for <charge conjugation>. To see the charge reversal explicitly, the conserved charge $Q_{\rm em}=\int_{\mathbf p}(a_{\mathbf p}^\dagger a_{\mathbf p}-b_{\mathbf p}^\dagger b_{\mathbf p})$ obeys $CQ_{\rm em}C^\dagger=-Q_{\rm em}$. The generator $Q$ in the question is distinct from this charge.
The two occupation terms in the <Hamiltonian> are exchanged, with the same <relativistic energy> $E_{\mathbf p}$, and therefore
$$
\boxed{CHC^\dagger=H.}
$$
In fact $[Q,H]=0$ because its mixing terms exchange equal-<energy> modes. With <normal ordering>, $:Q:=2\int_{\mathbf p}c_{-,\mathbf p}^\dagger c_{-,\mathbf p}$ for $c_-=(a-b)/\sqrt2$. Then $C=e^{i\pi N_-}$ fixes the <vacuum state> and satisfies $C^2=1$. The construction is a <bosonic species-swap generator>.
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