In a chemical reaction where sulphuric acid reacts with copper (II) carbonate to form copper(II) sulphate, water, and carbon dioxide, what is the color change of a blue litmus paper at the start and at the end of the reaction? (Assuming it is the same piece of litmus paper)

This question is about conjugate acids and bases. It could be answered at a number of different levels. Since no concentrations are given, I'm going to assume that a qualitative answer will be acceptable and you are not expected to analyze all of the equilibria and calculate concentrations of H_3O^+ and OH^- ions.
In this reaction, sulfuric acid reacts with carbonate ion, a weak base, in an acid-base neutralization reaction. Let's look at the start of the reaction first. I have not seen a question asked this way but will assume that "the start" means we are to compare the concentrations of H_3O^+ produced by the acid and OH^- produced by the base to determine whether the mixture is initially acidic or basic.
If the molar amounts of acid and base are different, the outcome will be determined by whether there is more acid or more base. It seems likely that this question wants you to consider the case in which the molar quantities are exactly equal, so that is what we will examine.
Sulfuric acid is a diprotic acid, so we can write two equations for its ionization in water. It is considered a "strong" acid, meaning ionization goes to completion, but this only applies to the first ionization; the dissociation of hydrogen sulfide ion goes to equilibrium. Thus, the first reaction has a one-way arrow, while the second has an equal sign representing an equilibrium arrow:
H_2SO_4 + H_2O -> H_3O^+ + HSO_4^-; K_a > 1
HSO_4^(-) + H_2O = H_3O^+ + SO_4^(2-); K_a = 1.0 xx 10^(-2)
I've included a link to the table of K_a values I used.
Carbonate ion CO_3^(2-) (from the copper (II) carbonate) is a weak base. It is the conjugate base of the acidic dissociation of hydrogen carbonate ion HCO_3^- , which is, itself, the conjugate base formed in the first dissociation of carbonic acid, H_2CO_3 .
The equilibria for carbonate ion, hydrogen carbonate ion, and carbonic acid can be written
CO_3^(2-) + H_2O = HCO_3^(-) + OH^-; K_b = 2.1 xx 10^(-4)
HCO_3^(-) + H_2O = H_2CO_3 + OH^-; K_b = 2.3 xx 10^(-8)
Here I have calculated the K_b values from the linked table of K_a values for the conjugate acids.
Qualitatively comparing values of the equilibrium constants, you can see that the K_a values for both ionizations of sulfuric acid are much greater than the K_b values for the reactions of carbonate and hydrogen carbonate ions with water. This means that at the time the reactants are mixed together, there will be a lot more H_3O^+ present than OH^- . These two ions will quickly come to equilibrium according to
2 H_2O = H_3O^+ + OH^-; K_w = 1.0 xx 10^(-14)
and the equilibrium will be dominated by H_3O^+ . Thus, according to our assumptions about the "start" of the reaction, the reaction mixture will be acidic. Blue litmus paper dipped in acidic solution will turn red.
What about at the end of the reaction? The question notes that the end products are copper sulfate, water, and carbon dioxide. A brief explanation of the carbon dioxide: aqueous carbonic acid (formed when carbonate ion, and then hydrogen carbonate ion, are protonated by the sulfuric acid) exists in equilibrium with dissolved carbon dioxide according to
H_2CO_3 (aq) = H_2O (l) + CO_2 (aq)
This is an equilibrium; it does not go to completion, so we might expect the final mixture to contain equilibrium quantities of every ion mentioned so far—except for one more reaction. There is a phase equilibrium between dissolved carbon dioxide and carbon dioxide gas
CO_2 (aq) = CO_2 (g) .
If the reaction vessel is open to the air, carbon dioxide gas will leave the system, pulling this equilibrium to the right, according to LeChatelier's Principle. Removal of CO_2 will shift every equilibrium in the system to the right. I infer from the statement of the problem that the "end" of the reaction is considered to be when all CO_2 is in the gas phase, all of the carbonate and hydrogen carbonate will be used up leaving only the products stated in the problem: copper (II) ions, sulfate ions, and carbon dioxide.
But we are not finished, because sulfate ion SO_4^(2-) is the conjugate base of the weak acid hydrogen sulfate ion HSO_4^- . It reacts with water according to:
SO_4^(2-) + H_2O = HSO_4^(-) + OH^-; K_b = 1.0 xx 10^(-12)
This a a weakly basic solution and is expected to turn red litmus paper blue. Recall the blue litmus used at the start of the reaction was expected to turn red; dipping in this solution would be expected to change it back.
https://depts.washington.edu/eooptic/links/acidstrength.html