$\text{a)}{5}^0={\left(-5\right)}^0=1$  

---

$\text{b)}-2^4<{\left(-2\right)}^4$  

bo

$-2^4<0$  

${\left(-2\right)}^4>0$  

---

$\text{c)}-5^3={\left(-5\right)}^3=-125$  

---

$\text{d)}\frac{2^3}{3}>{\left(\frac{2}{3}\right)}^3$  

bo

$\frac{2^3}{3}=\frac{8}{3}$  

${\left(\frac{2}{3}\right)}^3=\frac{2^3}{3^3}=\frac{8}{27}$  

$\frac{8}{3}>\frac{8}{27}$  

---

$\text{e)}\frac{-1}{4^3}<{\left(\frac{-1}{5}\right)}^3$  

bo

$\frac{-1}{4^3}=-\frac{1}{64}$  

${\left(\frac{-1}{5}\right)}^3=-\frac{1}{125}$  

$-\frac{1}{64}<-\frac{1}{125}$  

---

$\text{f)}\frac{2}{-8^{10}}<\frac{-2}{8^{11}}$  

bo obie liczby są ujemne i

$8^{10}<8^{11}$