Help with combustion analysis?

Yes, the given amount of mass that the respective "absorbers" gain is the mass of the H₂O and CO₂ that are formed as a result of the combustion. In other words :
Mass of H₂O formed in the combustion = Gain in mass of the H₂O absorber = 0.166 g
Mass of CO₂ formed in the combustion = Gain in mass of the CO₂ absorber = 0.403 g


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Solution (method 1) to the question :

Molar masses :
H = 1.0 g/mol
C = 12.0 g/mol
O = 16.0 g/mol
H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol

Mass of C in 0.165 g of valproic acid = Mass of C in CO₂ formed = (0.403 g) × (12.0/44.0) = 0.1099 g
Mass of H in 0.165 g of valproic acid = Mass of H in H₂O formed = (0.166 g) × (1.0×2/18.0) = 0.0184 g
Mass of O in 0.165 g of valproic acid = (0.165 - 0.1099 - 0.0184) g = 0.0367 g

Mole ratio C : H : O
= 0.1099/12.0 : 0.0184/1.0 : 0.0367/16
= 0.00916 : 0.0184 : 0.00229
= 0.00916/0.00229 : 0.0184/0.00229 : 0.00229/0.00229
= 4 : 8.03 : 1
≈ 4 : 8 : 1

Empirical formula of valproic acid = C₄H₈O

Let (C₄H₈O)n be the molecular formula of valproic acid.

Molar mass of valproic acid (in g/mol) :
(12.0×4 + 1.0×8 + 16.0) × n = 144
72.0 × n = 144
n = 2

Molecular formula of valproic acid = C₈H₁₆O₂


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Solution (method 2) to the question :

Molar masses :
H = 1.0 g/mol
C = 12.0 g/mol
O = 16.0 g/mol
H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol

Mass of C in 0.165 g of valproic acid = Mass of C in CO₂ formed = (0.403 g) × (12.0/44.0) = 0.1099 g
Mass of H in 0.165 g of valproic acid = Mass of H in H₂O formed = (0.166 g) × (1.0×2/18.0) = 0.0184 g
Mass of O in 0.165 g of valproic acid = (0.165 - 0.1099 - 0.0184) g = 0.0367 g

Let CxHyOz be the molecular formula of valproic acid.

x = 144 × (0.1099/0.165) / 12.0 = 7.99 ≈ 8
y = 144 × (0.0184/0.165) / 1.0 = 16.1 ≈ 16
z = 144 × (0.0367/0.165) / 16.0 = 2

Molecular formula of valproic acid = C₈H₁₆O₂