why is dH=dE at constant volume?
It isn't, and nowhere in the problem or answer is this implied.
First of all some definitions.
For a combustion reaction, the enthalpy change can be equated with the heat of combustion at constant pressure, whereas the internal energy is the heat of combustion at constant volume:
$$\Delta U = q_V ~~~~~\text{constant volume} \\ \Delta H = q_p ~~~~~ \text{constant pressure}$$
There is a simple integral form for the relation between internal energy and enthalpy that follows from the definition of the latter:
$$\begin{align} \Delta H &= \Delta (U+pV) \\ &= \Delta U + \Delta (pV) \end{align}$$
This relation always holds (it is a definition).
For this particular problem, the key to the next step is to use the ideal gas law, $pV=n_gRT$. This is allowed only if that equation of state holds for the substances in question (here carbon dioxide and oxygen). Inserting the ideal gas relation gives
$$\begin{align} \Delta H &= \Delta U + \Delta (n_gRT) \\ &= \Delta U + RT\Delta n_g\end{align}$$
where T is assumed constant. This is the expression used to find the solution. Clearly it shows that the change in enthalpy and internal energy are different.
You could derive this last expression slightly differently to emphasize the differences in the energy and enthalpy. The enthalpy change includes the heat input required to perform work against a constant pressure atmosphere and isothermally. In the case of an ideal gas, this heat input is
$$q_p-q_V = \Delta q = -w_{pV} = p\Delta V = p \Delta (n_g RT/p) = RT \Delta n_g$$
Note that for an ideal gas the enthalpy is a function of T only, so it's true then that a change in enthalpy is the same under constant V or p, provided you converted equal amounts of substance at same T. The same holds for the energy (since for an ideal gas it is also a function of T only).