A solid is a mixture of sodium carbonate, sodium bicarbonate and an inert salt…

Carbonate. The first stage converts all $\text{CO}_3^{2-}$ to $\text{HCO}_3^-$ with one $\text{H}^+$ each: $6.00 \times 0.100 = 0.60\ \text{mmol}$ in the portion.

Bicarbonate. The second stage turns every $\text{HCO}_3^-$ into $\text{CO}_2$. That is the bicarbonate formed from the carbonate ($0.60\ \text{mmol}$) plus the original bicarbonate. So the original bicarbonate is $8.00 \times 0.100 - 0.60 = 0.20\ \text{mmol}$.

(a), (b), (c) The portion is $1/10$ of the sample. Carbonate: $0.60 \times 10 = 6.0\ \text{mmol}$, which is $0.636\ \text{g}$ of $\text{Na}_2\text{CO}_3$ (106.0) and so $31.8\%$. Bicarbonate: $0.20 \times 10 = 2.0\ \text{mmol}$, which is $0.168\ \text{g}$ of $\text{NaHCO}_3$ (84.0) and so $8.4\%$. The inert salt is $1.196\ \text{g}$, which is $59.8\%$.

(d) The $1.000\ \text{g}$ portion holds $3.00\ \text{mmol}$ of carbonate and $1.00\ \text{mmol}$ of bicarbonate. It uses $2(3.00) + 1.00 = 7.00\ \text{mmol}$ of $\text{H}^+$. The acid added is $50.0 \times 0.200 = 10.00\ \text{mmol}$, so $3.00\ \text{mmol}$ is left, and the NaOH needed is $3.00 / 0.100 = 30.0\ \text{mL}$.

(e) $\text{CO}_2$ is $3.00 + 1.00 = 4.00\ \text{mmol}$, so the volume is $4.00 \times 22.4 = 89.6\ \text{mL}$.