Preparing aqueous solution of ethanol You need to prepare $\pu{250.0 mL}$ of an aqueous solution of ethanol in the concentration will be $\pu{0.500 mol L-1}$. You have $\pu{100 mL}$ of a commercial solution of ethanol in water $(35.0\% \,V/V)$ available. The density of ethanol (pure) under these conditions is $\pu{0.7893 g mL-1}$ and that of water is $\pu{0.9970 g / mL}$.
1.What volume (in $\mathrm{mL}$) will you need to withdraw to prepare this solution?
2.What is the mole fraction of ethanol in the resulting solution?

Answer to (1)
We can prepare $\pu{250 mL}$ of $\pu{0.5 mol L-1}$ solution of ethanol by diluting $\pu{20.85 mL}$ of $35 \%$ $\ce{C2H5OH}$ (density $\pu{0.7893 g/mL}$) with distilled water to a volume of $\pu{250 mL}$.
The dilution is therefore made by adding $\pu{229.15 mL}$ of water.
However, the volume of a liquid obtained by mixing measured volumes of different solutions is not always precisely the sum of the component volumes.
Calculation
Source (B): $35 \%$ solution of $\ce{C2H5OH}$ (density $\pu{0.7893 g/mL}$)
The molar concentration of a solution of a chemical species A is the number of moles of that species contained in one litre of the solution (not in one litre of the solvent).
$$C(A) = \pu{0.5 mol/L}$$
$$\text{Molar concentration} = \frac{\text{Density} \cdot \text{Weight percent} }{ 100\cdot \text{Molecular weight}}$$
$$C(B) = \frac{d(B) \cdot w(B)}{100 \cdot M(B)}$$
$$C(B) = \frac{\pu{789.3 g/L} \times 35 \%}{100 \times \pu{46.0684 g/mol}}$$
$$C(B) = \pu{5.99662675499909 mol/L}$$
The number of moles of solute in the diluted solution must equal the number of moles in the concentrated reagent.
\begin{align}
V(A) \cdot C(A) &amp;= V(B) \cdot C(B) \\ 
V(B) &amp;= \frac{V(A) \cdot C(A)}{C(B)} \\
V(B) &amp;= \frac{\pu{0.25 L} \times \pu{0.5 mol/L}}{\pu{5.99662675499909 mol/L}} \\
V(B) &amp;= \pu{2.08450525782339 \times 10^{-2} L} \\
V(B) &amp;= \pu{20.85 mL} \\
\end{align}
Answer to (2)
We have
\begin{align}
C(B)  &amp;= \frac{\pu{0.125029667842 mol}}{\pu{20.85 mL}} \\
&amp;=\pu{5.99662675499909 mol/L} 
\end{align}
Now we have to add $\pu{229.15 mL}$ of water to $\pu{20.85 mL}$ of commercial solution of ethanol to make it $\pu{250 mL}$.
$\pu{229.15 mL}$ of water has a mass of $\pu{229.15 g}$. This means that there are $\pu{12.7197 mol}$ water because the molar mass of water is $\pu{18.0153 g}$
\begin{align}
\text{Mole fraction of ethanol}&amp;= \frac{\text{Number of moles of ethanol}}{\text{Number of moles of solution}} \\
&amp;=\frac{\pu{0.125029667842 mol}}{(\pu{0.125029667842 mol} +\pu{12.7197 mol}}\\ &amp;
= \pu{0.009734 mol}
\end{align}

Are my answers to both questions correct? If they are incorrect, where am I going wrong?
@ Dhamnekar. Your calculations are correct, provided your assumption (that the volumes are additive) is correct. Nevertheless note that the last value ($x = 0.0097$) has no unit : it is not in moles. And we must also admit that, due to your assumption, x is probably simply $x = 0.010$.
Anyway, next time try to trust your calculations yourself. We are not here to verify correct calculations. We are here to HELP THOSE WHO DO NOT SUCCEED in solving their problems.