\(\begin{array}{l}
a)\\
Dat\,n_{Fe}=1(mol);n_{H_2SO_4(du)}=x(mol)\\
2Fe+6H_2SO_4\to Fe_2(SO_4)_3+3SO_2+6H_2O\\
Theo\,PT:\,n_{H_2SO_4(pu)}=3;n_{Fe_2(SO_4)_3}=0,5;n_{SO_2}=1,5\\
\to m_{dd\,H_2SO_4(bđ)}=\dfrac{(3+x).98}{78,4\%}=125(3+x)(g)\\
\to m_{dd\,spu}=1.56+125(3+x)-1,5.64=335+125x(g)\\
C\%_{Fe_2(SO_4)_3}=C\%_{H_2SO_4(du)}\\
\to \dfrac{0,5.400}{335+125x}.100=\dfrac{98x}{335+125x}.100\\
\to 98x=200\to x=\dfrac{100}{49}(mol)\\
\to C\%_{Fe_2(SO_4)_3}=C\%_{H_2SO_4(du)}\approx 34\%\\
b)\\
n_{Fe_2(SO_4)_3}=\dfrac{50.34\%}{400}=0,0425(mol)\\
n_{H_2SO_4(du)}=\dfrac{50.34\%}{98}=\dfrac{17}{98}(mol)\\
2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\
Fe_2(SO_4)_3+6NaOH\to 2Fe(OH)_3+3Na_2SO_4\\
Theo\,PT:\,\sum n_{NaOH}=2.\dfrac{17}{98}+6.0,0425\approx 0,6(mol)\\
\to V_{dd\,NaOH}=\dfrac{0,6}{2}=0,3(l)=300(ml)
\end{array}\)