What is the indicator used in Argentometric titrations?

$0.4 \ g$ mixture of $NaOH, Na_{2}CO_{3}$ and some inert impurities was first titrated with $\frac{N}{10} \ HCl$ using phenolphthalein as an indicator,$17.5 \ mL$ of $HCl$ was required at the end point. After this,methyl orange was added and titrated. $1.5 \ mL$ of same $HCl$ was required for the next end point. The weight percentage of $Na_{2}CO_{3}$ in the mixture is .......... (Rounded-off to the nearest integer)

Given below are two statements:
Statement $I$: In the oxalic acid vs $KMnO_4$ (in the presence of dil. $H_2SO_4$) titration,the solution needs to be heated initially to $60^{\circ}C$,but no heating is required in Ferrous Ammonium Sulphate $(\text{FAS})$ vs $KMnO_4$ titration (in the presence of dil. $H_2SO_4$).
Statement $II$: In oxalic acid vs $KMnO_4$ titration,the initial formation of $MnSO_4$ takes place at high temperature,which then acts as a catalyst for the further reaction. In the case of $\text{FAS}$ vs $KMnO_4$,heating oxidizes $Fe^{2+}$ into $Fe^{3+}$ by oxygen of air and error may be introduced in the experiment.
In the light of the above statements,choose the correct answer from the options given below:

$20 \, mL$ of $0.02 \, M$ hypo solution is used for the titration of $10 \, mL$ of copper sulphate solution,in the presence of excess of $KI$ using starch as an indicator. The molarity of $Cu^{2+}$ is found to be $\times 10^{-2} \, M$ [nearest integer]. Given: $2 Cu^{2+} + 4 I^{-} \rightarrow Cu_{2}I_{2} + I_{2}$ and $I_{2} + 2 S_{2}O_{3}^{2-} \rightarrow 2 I^{-} + S_{4}O_{6}^{2-}$.

Consider the titration of $NaOH$ solution versus $1.25\, M$ oxalic acid solution. At the end point,the following burette readings were obtained:
$(i)$ $4.5\, mL$ $\quad (ii)$ $4.5\, mL$ $\quad (iii)$ $4.4\, mL$
$(iv)$ $4.4\, mL$ $\quad (v)$ $4.4\, mL$
If the volume of oxalic acid taken was $10.0\, mL$,then the molarity of the $NaOH$ solution is .... $M$ (Rounded-off to the nearest integer).

$10 \, mL$ of concentrated $HCl$ is diluted to $1 \, L$. If $20 \, mL$ of this diluted solution is completely neutralized by $25 \, mL$ of $0.1 \, N$ sodium hydroxide solution,then the normality of the original concentrated hydrochloric acid is: