a,
Gọi x, y là số mol Fe, Cu.
$\Rightarrow 56x+64y=17,12$ (1)
$Fe+4HNO_3\to Fe(NO_3)_3+NO+2H_2O$
$3Cu+8HNO_3\to 3Cu(NO_3)_2+2NO+4H_2O$
$n_{NO}=\dfrac{4,928}{22,4}=0,22(mol)$
$\Rightarrow x+\dfrac{2}{3}y=0,22$ (2)
$(1)(2)\Rightarrow x=0,1; y=0,18$
$m_{Fe}=56x=5,6g$
$m_{Cu}=64y=11,52g$
b,
$n_{HNO_3}=4x+\dfrac{8}{3}y=0,88(mol)$
$\to C_{M_{HNO_3}}=\dfrac{0,88}{2}=0,44M$