Решение.

Применяем свойства степеней:  

[tex]a^{-1}=\dfrac{1}{a}\ ,\ \ a^{-n}=\dfrac{1}{a^{n}}\ \ ,\ \ \dfrac{a^{n}}{a^{m}}=a^{n-m}\ ,\ a^0=1[/tex]  .

[tex]\displaystyle \bf \Big(\frac{1}{5}\Big)^{-1}\cdot 25^{-1}-\Big(-\frac{1}{4}\Big)^{-3}+(-0,4)^{-4}\cdot (-0,4)^5-(10^3)^0=\\\\\\=5\cdot \frac{1}{25}-\frac{1}{\Big(-\dfrac{1}{4}\Big)^3}+\frac{1}{0,4^4}\cdot \Big(-\frac{4}{10}\Big)^5-1=\\\\\\=\frac{1}{5}+4^3+\Big(\frac{10}{4}\Big)^4\cdot \Big(-\frac{4}{10}\Big)^5-1=\\\\\\=\frac{1}{5}+64-\Big(\frac{5}{2}\Big)^4\cdot \Big(\frac{2}{5}\Big)^5-1=\\\\\\=\frac{1}{5}+64-\frac{5^4\cdot 2^5}{2^4\cdot 5^5}-1=[/tex]

[tex]\displaystyle \bf =\frac{1}{5}+64-\frac{2}{5}-1=63-\frac{1}{5}=\frac{314}{5}=62\frac{4}{5} =62,8[/tex]