How do we approach this question?
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a) To have maximum precipitation we need the salt with minimum molar solubility.Here the Ksp of the SrCO₃ is minimum and it is the best salt.To be sure you can calculate molar solubility of Li₂CO₃ and SrCO₃ because they have different Ksp expression.
Li₂CO₃  = 4s³ =8.15 * 10^-4
SrCO₃   = s² = 5.60 *10^-10
b)For the SrCO₃ that contains the cation chosen in part (a), determine the concentration of each ion of that compound in solution at equilibrium.
SrCO₃(s) ⇌ Sr2⁺(aq) + CO₃^2-(aq)
Ksp = [Sr²⁺][CO₃^2-]
5.60 × 10^-10 = (x)(x)
x = 2.37 × 10^-5 M = [Sr²⁺] = [CO₃^2-]
c)Student must ensure excess of nitrate solution to ensure the precipitaions of all carbonate ions from the given sample.

d) Convert g SrCO₃(s)  to moles SrCO₃(s) using molar mass ratio.

e) Convert moles SrCO₃(s) to moles of crabonate ( 1mole SrCO₃(s)  has one mole CO₃^2-) and then grams of carbonate using molar mass of CO₃^2- .
f) Mass of CO₃^2- calculated in e part divided by total mass of sample(1.89 g) given in the satrting of the question multiplied by 100%.
g) Now you know mass the CO₃^2- in orginal sample and moles CO₃^2- also.Subtract mass of carbonate from 1.89 g to get the mass of metal.
Now moles of metal = 2 moles of carbonate as given by the type of salt.
Li₂CO₃ ,Na₂CO₃ and K₂CO₃ .
It means you know the mass and moles so get the molar mass of the element to know its identity.
moles = mass / molar mass.