Anodisation of aluminium in sulfuric acid - Electrolysis reactions
As the experiment is described, in the solution we have : 
$$H^{+}, H_{2}O, Al_{(s)}, SO_{4}^{2-}, OH^{-}$$
Taking that into account and seeing how the environment is acidic ( $pH&lt;7$ ), we can see that the reaction of reduction reaction that can take place is $H^{+}/H_{2}$
Following the same kind of reasoning, reaction $(1)$ in the anode is caused by the odydation of the aluminium plate when placed in an acidic environment, $O^{-2}$ is present because of the oxygen in the atmosphere and $OH^{-}$ exists in the solution anyway(water autoionization).

As far as the second question is concerned, notice that the aluminium foil in the anode does not participate in any reaction. It is the $H^{+}$ that is reduced, while the aluminium foil serves as a conductor for the current. Knowing that, we could replace the aluminium foil with either $Cu_{(s)}$ or $Pb_{(s)}$, that may also be used as conductors ($Cu^{2+}, Pb^{2+}$ will not be present in the solution and we will have the same reduction reaction in the anode), no matter their standard electrode potentials.

Finally, it is interesting to point out that in this experiment the anode and cathode are in the same solution which is used at the electrolytic reactions (acidic environment ) and at the same time plays the role of the electrolytic bridge constantly providing $H^{+}$ and keeping the solution electrically neutral.