$1)N_xO_y\Rightarrow \begin{cases}\frac{14x}{14x+16y}=30,43\% \\ 14x+16y=29.1,59 \end{cases}\Rightarrow \begin{cases}x=1 \\ y=2 \end{cases}\Rightarrow NO_2$$2)$
$Cu+4HNO_3\rightarrow Cu(NO_3)_2+2NO_2\uparrow+2H_2O$
$\Rightarrow m_{dd HNO_3}=\frac{63.100.(2n_{NO_2})}{40}=\frac{63.100.2.1.(134+273)}{40.22,4.273}=20,965 gam$
$3)$
$2NO_2\rightarrow N_2O_4$
$\begin{cases}NO_2:x mol \\ N_2O_4:y mol \end{cases}\Rightarrow \frac{46x+92y}{x+y}=29.1,752\Rightarrow x:y=9:1$
$a)\Rightarrow \begin{cases}\%NO_2=90\% \\ \%N_2O_4=10\% \end{cases}$
$b)\Rightarrow \%(A\rightarrow B)=\frac{2y.100}{x+2y}=18,18\%$
$4)$
$N_2O_4\overset{134^oC}{\rightarrow}2NO_2$
$3NO_2+H_2O\rightarrow 2HNO_3+NO\uparrow$
$n_{NO_2}=\frac{5(25+273)(90\%+2.10\%)}{22,4.273}=0,268 mol$
$\Rightarrow [HNO_3]=\frac{0,268.2}{3.5}=0,035733 M$
$\%(A\rightarrow D)=\frac{2.100}{3}=66,67\%$