$a.$

$Al+4HNO_3 \to Al(NO_3)_3 +NO+2H_2O$

$Fe+4HNO_3 \to Fe(NO_3)_3+NO+2H_2O$

$b.$

$n_{NO}=0.3(mol)$

$n_{Al}=a(mol);n_{Fe}=b(mol)$

`->`$27a+56b=13.9$ $(1)$

Bảo toàn $e:$

$3n_{Al}+3n_{Fe}=3n_{NO}$

`=>`$a+b=0.3$ $(2)$

Từ  $(1)$ và $(2)$ suy ra: $a=0.1;b=0.2$

`=>`$\%m_{Al}=\dfrac{0.1 \times 27}{13.9} \times 100\%=19.42\%$

`=>`$\%m_{Fe}=(100-19.42)\%=80.58\%$

$c.$

$2Fe(NO_3)_3+Fe \to 3Fe(NO_3)_2$