Isolating barium(II), copper(II) and zinc(II) from aqueous solution
Once $\ce{Ba^2+}$ is eliminated as poorly soluble barium(II) sulfate, there is a solution of soluble copper(II) and zinc(II) sulfates.
Considering this has been done by careful/gradual addition of sulfuric acid, and the fact that both remaining sulfates undergo hydrolysis, the solution is going to be slightly acidic — this is one reason for sodium hydroxide "added in excess".

Another reason, as both Ivan and Waylander pointed out in the comments, is the difference between reactivity of copper(II) and zinc(II) in alkali medium.
Long story short, you are probably supposed to know the specific properties of the common transition metals cations as the reduction potential isn't really helpful here.

Copper(II)

Copper(II) with $\ce{NaOH}$ first forms colloidal copper(II) hydroxide; the gel is then solidifies to light-blue amorphous substance upon subsequent addition of hydroxide:

$$\ce{CuSO4(aq) + 2 NaOH(aq) -&gt; Cu(OH)2(s) + Na2SO4(aq)}$$

$\ce{Cu(OH)2(s)}$ appears to be precipitate b.
Sodium tetrahydroxocuprate(II) $\ce{Na2[Cu(OH)4]}$ doesn't form here: it's a very moisture-sensitive and in the air quickly turns dark-brown:

$$\ce{Na2[Cu(OH)4](aq) -&gt; CuO(s) + H2O(l) + 2 Na(OH)(aq)}$$

Zinc(II)

Zinc(II), on the other hand, with $\ce{NaOH}$ first forms white precipitate of zinc(II) hydroxide:

$$\ce{\ce{ZnSO4(aq) + 2 NaOH(aq) -&gt; Zn(OH)2(s) + Na2SO4(aq)}}$$

which is then dissolved again forming colorless soluble sodium terahydroxozincate(II) (filtrate c):

$$\ce{Zn(OH)2(s) + 2 NaOH(aq) -&gt; Na2[Zn(OH)4](aq)}$$

Excess of $\ce{NaOH}$ is also necessary to keep $\ce{Na2[Zn(OH)4]}$ in solution: upon diluting (lowering pH, adding water/acid), terahydroxozincate(II) decomposes to zinc(II) hydroxide.

Note: chemical reactions are adapted from [1].

References


R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)