So, the aluminium cation acts as a weak acid, its acid dissociation constant, #K_a#, being equal to #1 * 10^(-5)#.
You can use the value of its acid dissociation constant and the molarity of the solution to determine the concentration of hydronium ions, #H_3O^(+)#, present in solution.
For a weak acid, you have
#[H_3O^(+)] = sqrt(K_a * C)#
#[H_3O^(+)] = sqrt( 1 * 10^(-5) * 2.5) = 0.005#
Now that you have the concentration of the hydronium ions, you can determine the concentration of the hydroxide ions by
#[OH^(-)] = 10^(-14)/([H_3O^(+)])#
#[OH^(-)] = 10^(-14)/(5 * 10^(-3)) = 2 * 10^(-12)#
Now, in order to determine whether or not a precipitate will form, you need the solubility product constant, #K_(sp)#, for aluminium hydroxide, which is listed as being equal to #1.3 * 10^(-33)#.
#Al(OH)_(3(s)) rightleftharpoons Al_text((aq])^(3+) + color(red)(3)OH_((aq))^(-)#
By definition, #K_(sp)# is
#K_(sp) = [Al^(3+)] * [OH^(-)]^(color(red)(3))#
The minimum concentration of hydroxide ions needed to precipitate aluminium hydroxide, given that the concentration of #Al^(3+)# is 2.5 M, will be
#[OH^(-)] = root(3)(K_(sp)/([Al^(3+)])) = root(3)((1.3 * 10^(-33))/2.5) = 8.0 * 10^(-12)#
Compare this value to the one you've calculated for #[OH^(-)]#, and you'll see that you don't have enough hydroxide ions present to form a precipitate
#2 * 10^(-12) < 8 * 10^(-12)# #=># NO precipitate is formed.