$$\begin{gathered} \mathrm{a}, \\ \mathrm{Ag}+2{\mathrm{HNO}}_3\to {\mathrm{AgNO}}_3+{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{a} \to \mathrm{a} \\ \mathrm{Cu}+4{\mathrm{HNO}}_3\to \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+2{\mathrm{NO}}_2+2{\mathrm{H}}_2\mathrm{O} \\ \mathrm{b} \to 2\mathrm{b} \\ \mathrm{b}, {\mathrm{n}}_{{\mathrm{NO}}_2}=0,04 \mathrm{mol} \\ => \mathrm{a}+2\mathrm{b}= 0,04 \left(1\right) \\ \mathrm{Theo} \mathrm{ài} 108\mathrm{a}+64\mathrm{b}= 2,04 \left(2\right) \\ (1,2)=> \mathrm{a}=0,01; \mathrm{b}=0,015 \\ {\mathrm{m}}_{\mathrm{Ag}}=0,01.108=1,08 \mathrm{g} \\ {\mathrm{m}}_{\mathrm{Cu}}=0,015.64=0,96 \mathrm{g} \\ \mathrm{c}, \\ \mathrm{Co}: {\mathrm{m}}_{{\mathrm{AgNO}}_3}+{\mathrm{m}}_{\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2}=4,52 \mathrm{g} \\ \mathrm{Sau} \mathrm{khi} \mathrm{đun} \mathrm{nóng} \mathrm{thu} \mathrm{được} \mathrm{Ag}; \mathrm{CuO} \\ \mathrm{Theo} \mathrm{BTNT}: {\mathrm{n}}_{\mathrm{Ag}}=0,01 \mathrm{mol} \\ {\mathrm{n}}_{\mathrm{CuO}}={\mathrm{n}}_{\mathrm{Cu}}=0,015 \mathrm{mol} \\ =>\hspace{0.17em}{\mathrm{m}}_{\mathrm{giảm}}= {\mathrm{m}}_{{\mathrm{AgNO}}_3}+{\mathrm{m}}_{\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2}-{\mathrm{m}}_{\mathrm{Ag}}-{\mathrm{m}}_{\mathrm{CuO}}=2,24 \mathrm{g} \end{gathered}$$.

$$\begin{gathered} n_{N{O}_2}=0,04 mol \\ a, Ag+2HN{O}_3->AgN{O}_3+N{O}_2+H_{2}O \\ x------> x x \left(mol\right) \\ Cu+4HN{O}_3->Cu{\left(N{O}_3\right)}_2+2N{O}_2+2H_{2}O \\ y----->y 2y \left(mol\right) \\ b, Ta có: \left\{\begin{array}{l}x+2y=0,04\\ 108x+64y=2,04\end{array}<=>\left\{\begin{array}{l}x=0,01\\ y=0,015\end{array}\right.\right. \\ =>m_{Ag}=0,01.108=1,08g \\ {m}_{Cu}=2,04-1,08=0,96g \\ Vậy hh có 1,08g Ag và 0,96g Cu \end{gathered}$$

$$\begin{gathered} c, 2Cu{\left(N{O}_3\right)}_2\stackrel{to}{\to }2CuO+4N{O}_2+O_2 \\ 0,015------>0,03 0,0075 \left(mol\right) \\ 2AgN{O}_3\stackrel{to}{\to }2Ag+2N{O}_2+O_2 \\ 0,01------>0,01 0,005 \left(mol\right) \\ {m}_{giảm}=m_{N{O}_2}+m_{O_2}=2,24g \\ Vậy m_{giảm}=2,24g \end{gathered}$$.