Given below are two statements:

Statement I: First I.E order: Na > Mg > Al

Statement II: Third I.E order: Mg > Al > Na

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To determine the correctness of the given statements, let us analyze the first and third ionization energies ($I.E.$) of Sodium ($Na$), Magnesium ($Mg$), and Aluminum ($Al$).

1. Analysis of Statement I (First Ionization Energy, $I.E_1$):
The electronic configurations of the neutral gaseous atoms are:
• $Na:1s^{2}2s^{2}2p^{6}3s^1$
• $Mg:1s^{2}2s^{2}2p^{6}3s^2$
• $Al:1s^{2}2s^{2}2p^{6}3s^{2}3p^1$

• Sodium ($Na$) has only one electron in its valence shell ($3s^1$), which can be removed very easily. Therefore, it has the lowest $I.E_1$ among the three elements.
• Magnesium ($Mg$) has a fully-filled, stable valence shell configuration ($3s^2$).
• Aluminum ($Al$) has the configuration $3s^{2}3p^1$. The electron is to be removed from the $3p$ subshell, which is higher in energy and experiences more shielding than the $3s$ electrons in $Mg$. Consequently, it is easier to remove an electron from $Al$ than from $Mg$.

Thus, the correct first ionization energy order is:
$Na<Al<Mg$
Therefore, Statement I ($Na>Mg>Al$) is incorrect.

2. Analysis of Statement II (Third Ionization Energy, $I.E_3$):
To determine the third ionization energy, we look at the electronic configurations of the divalent cations ($M^{2+}$), from which the third electron is removed:
• $N{a}^{2+}:1s^{2}2s^{2}2p^5$
• $M{g}^{2+}:1s^{2}2s^{2}2p^6$ (Noble gas configuration)
• $A{l}^{2+}:1s^{2}2s^{2}2p^{6}3s^1$

• For $M{g}^{2+}$, the third electron must be removed from a highly stable, closed-shell noble gas configuration ($2p^6$). This requires an exceptionally large amount of energy. Thus, $Mg$ has the highest $I.E_3$.
• For $N{a}^{2+}$, the electron is removed from the $2p$ subshell (principle quantum number $n=2$).
• For $A{l}^{2+}$, the electron is removed from the valence $3s$ subshell (principle quantum number $n=3$), which is further from the nucleus and shielded by the inner shells. Hence, it is much easier to remove than the $2p$ electron of $N{a}^{2+}$.

Thus, the correct third ionization energy order is:
$Mg>Na>Al$
Therefore, Statement II ($Mg>Al>Na$) is also incorrect.

Conclusion:
Since both Statement I and Statement II are incorrect, the correct option is "Both Statement I and Statement II are incorrect".