How to calculate the concentrations of the species in the carbonate equilibrium from a titration with hydrochloric acid?
You have correctly identified the amount of substance $\ce{HCl}$ which were necessary to neutralize all $\ce{HCO3-}$ and $\ce{CO3^2-}$ in the sample. With the given sample volume of $\pu{10 cm^3}$, we can calculate the concentration of both ions in the sample:
$$
\frac{\pu{4.4E-3 mol}}{\pu{10 cm^3}}
= \pu{4.4E-4 mol//cm^3}
= [\ce{HCO3-}]+[\ce{CO3^{2-}}]
$$

Using the given ratio of $0.958$, we can express $[\ce{CO3^{2-}}]$ in terms of $[\ce{HCO3^{-}}]$:
\begin{gather}
[\ce{CO3^2-}] 
  = 0.958\times[\ce{HCO3-}]\\
\pu{4.4E-4 mol//cm^3} 
  = [\ce{HCO3-}] + [\ce{CO3^2-}] 
  = [\ce{HCO3-}] + 0.958\times[\ce{HCO3-}] 
  = 1.958[\ce{HCO3-}]\\
[\ce{HCO3-}]
  = \pu{2.25E-4 mol//cm3}
\end{gather}

We then use the calculated hydrogen carbonate concentration and the ratio to find $[\ce{CO3^2-}]$:
$$[\ce{CO3^2-}] = 0.958\times[\ce{HCO3-}] = \pu{2.15E-4 mol//cm3}$$