Hi I am not 100% sure if this is right but this is what I think
so the decomposition of 2NaHCO3 -> Na2CO3 + H2O + CO2
so if I let the number of moles of Na2CO3 produced as x, then the equation will be
2xNaHCO3 -> xNa2CO3 + xH2O + xCO2
so xCO2 + xH2O = 10-8.708
62x = 1.292
x = 0.02084 moles
so mass of NaHCO3 = 2x0.02084x(23+1+12+16x3) = 3.501g
% of NaHCO3 = 3.501/10 x 100% = 35%
% of Na2CO3 = 1-35% = 65%