[tex]\displaystyle 1)\frac{2^3*4^-^2*8}{2^2*16}=\frac{2^3*2^-^4*2^3}{2^2*16}=\frac{2^2}{2^2*16}=\frac{1}{16}=0,0625\\ \\2)\sqrt{113^2-112^2}+(\sqrt{7}+3)(\sqrt{7}-3)=\sqrt{(113-112)(113+112)}+7-9=\\\\\sqrt{1*225}+7-9=\sqrt{225}+7-9=15+7-9=13[/tex]