Comparing Electron affinity and electron gain enthalpy at 0 K
In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs?

Successive Electron Gain Enthalpies
After the addition of one electron atom becomes negatively charged and second electron is to be added to a negatively charged ion. Thus, due to electrostatic repulsion energy is always required for addition of second electron. First electron gain enthalpy is always more negative than second electron gain enthalpy (which is essentially positive). Author made an ambiguous statement, as we can't just compare the magnitudes. For more clarity see:
$$\ce{S_{(g)} + e- -&gt; S- _{(g)}}             \Delta_{eg} H^- _1 = \pu{-200kJ mol^{-1}}$$
$$\ce{S- _{(g)} + e- -&gt; S^{2-}_{(g)}}             \Delta_{eg} H^- _2 = \pu{600kJ mol^{-1}}$$.

Does the same comparision hold with Electron Affinity at 0K?

Generally, if energy is released when an electron is added to an atom, electron affinity is taken as positive, contrary to thermodynamic convention. That is, if $\Delta_{eg} H^-$ is negative EA is positive (since temperature mentioned is absolute zero).
At absolute zero, $\Delta_{eg} H^- = \pu{-EA}$. Thus if $\Delta_{eg} H^- _1$ is negative, EA is positive and successive EAs will be increasingly negative. Here, you can say first electron affinity is greater than second (actual numbers with sign).

For example, Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?

First, $\ce{Be}$ has negative electron affinity ($\pu{-0.5 eV atom^{-1}}$) and $\ce{F}$ has positive electron affinity ($\pu{+3.4 eV atom^{-1}}$). So, fluorine has more electron affinity than beryllium, consider the actual numbers with signs).

How to compare their Electron gain enthalpies also at 0K?

Since, at 0K, $\Delta_{eg} H^- = \pu{-EA}$. Thus, first electron enthalpy of fluorine is more negative than that of beryllium. Strictly speaking, it's value is less than that of beryllium.