Identifying the gas given ratio of its specific heats
I'd go about such a question looking at the easiest thing first: ideal gases. So, what do you know about the relations between $C_{p}$ and $C_{v}$ (both being the molar heat capacities at constant pressure and constant volume, respectively) for ideal gases? For one thing, it is well known that

\begin{equation}
  C_{p} - C_{v} = R
\end{equation}

with $R$ is the universal gas constant. You can rearrange this equation for the ratio you are after:

\begin{equation}
  \frac{C_{p} - C_{v}}{C_{v}} = \frac{R}{C_{v}} \\
  \frac{C_{p}}{C_{v}} - 1 = \frac{R}{C_{v}} \\
  \frac{C_{p}}{C_{v}} = \frac{R}{C_{v}} + 1 \ .
\end{equation}

Now, you need to find $C_{v}$. Here, the degrees of freedom that you mentioned come into the game, since the equation that helps us out is the equipartition law, stating that each excited quadratic degree of freedom contributes $\frac{1}{2} R T$ to the molar internal energy $U$

\begin{equation}
  U = \frac{1}{2} R T (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}})
\end{equation}

with $f_{\text{trans}}$, $f_{\text{rot}}$ and $f_{\text{vib}}$ being the quadratic degrees of freedom for translation, rotation and vibration respectively. It is important to note that only excited degrees of freedom contribute to $U$. Which ones are active depends heavily on temperatur. Vibrations are usually not excited at room temperatur (there are exceptions, e.g. iodine) - they usually need temperatures in the order of $1000 \, \text{K}$ but that varies greatly. Rotations are much easier to excite - I know of no example where they aren't excited at room temperature. Translations take very little energy to excite and can always be considered active. If all degrees of freedom are active you get: For linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 2$ and $f_{\text{vib}} = 3N - 5$, where $N$ is the number of atoms in the molecule. For non-linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 3$ and $f_{\text{vib}} = 3N - 6$.
Since $C_{v} = \frac{dU}{dT}$ you get

\begin{equation}
  C_{v} = \frac{1}{2} R (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}}) \ .
\end{equation}

Substituting this into the equation for the ratio between $C_{p}$ and $C_{v}$ yields

\begin{equation}
  \frac{C_{p}}{C_{v}} = \frac{2}{(f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}})} + 1 \ .
\end{equation}

Now, you only have to know how many degrees of freedom each of the molecules has and plug them into the equation.