$1.A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{100.103}$
⇔ $\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}$
⇔ $\dfrac{3}{2}A=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}$
⇔ $\dfrac{3}{2}A=1-\dfrac{1}{103}$
⇔ $\dfrac{3}{2}A=\dfrac{102}{103}$
⇔ $A=\dfrac{68}{103}$
$3.C=\dfrac{4^8.25^8-125^5.8^6}{16^5.25^8+32^3.125^5}$
$=\dfrac{(2^{2})^8.(5^{2})^8-(5^{3})^5.(2^{3})^6}{(2^{4})^5.(5^2)^8+(2^5)^3.(5^3)^5}$
$=\dfrac{2^{16}.5^{16}-5^{15}.2^{18}}{2^{20}.5^{16}+2^{15}.5^{15}}$
$=\dfrac{2^{16}.5^{15}(5-2^2)}{2^{15}.5^{15}(2^5.5+1)}$
$=\dfrac{2(5-4)}{32.5+1}$
$=\dfrac{2}{161}$
