$a)$  

$4^3=4\cdot 4\cdot 4=64$  

 

${\left(-12\right)}^2=\left(-12\right)\cdot \left(-12\right)=144$  

 

${\left(\frac{1}{3}\right)}^4=\frac{1^4}{3^4}=\frac{1}{81}$  

 

${\left(-0,5\right)}^3=\left(-0,5\right)\cdot \left(-0,5\right)\cdot \left(-0,5\right)=-0,125$  

 

${\left(-3\frac{1}{3}\right)}^2={\left(-\frac{10}{3}\right)}^2=\frac{{\left(-10\right)}^2}{3^2}=\frac{100}{9}=\frac{99+1}{9}=11\frac{1}{9}$  

 

$1^{20}=\underset{20\text{razy mamy liczbę }1}{\underbrace{1\cdot 1\cdot 1\cdot \dots \cdot 1}}=1$  

 

${\left(-10\right)}^5=\left(-10\right)\cdot \left(-10\right)\cdot \left(-10\right)\cdot \left(-10\right)\cdot \left(-10\right)=-100000$  

 

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$b)$  

$\frac{3^2}{27}=\frac{3\cdot 3}{27}=\frac{9^{:9}}{{27}_{:9}}=\frac{1}{3}$  

 

$-\frac{3^2}{4}=-\frac{3\cdot 3}{4}=-\frac{9}{4}=-\frac{8+1}{4}=-2\frac{1}{4}$  

 

$-{\left(-\frac{5}{6}\right)}^3=-\left(\frac{{\left(-5\right)}^3}{6^3}\right)=-\left(\frac{-125}{216}\right)=\frac{125}{216}$  

 

${\left(0,1\right)}^3=0,1\cdot 0,1\cdot 0,1=0,001$  

 

$-{\left(0,2\right)}^2=-\left(0,2\cdot 0,2\right)=-0,04$  

 

${\left(-0,2\right)}^2=\left(-0,2\right)\cdot \left(-0,2\right)=0,04$  

 

${\left(-10\right)}^0=1$