$a\text{)}f\left(x\right)=2^x,x_1=-2,x_2=0,x_3={\log }_{2}5$  

Zatem:

$f\left(x_1\right)=f\left(-2\right)=2^{-2}={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$  

$f\left(x_2\right)=f\left(0\right)=2^0=1$  

$f\left(x_3\right)=f\left({\log }_{2}5\right)=2^{{\log }_{2}5}=5$  

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$b\text{)}f\left(x\right)=3^x,x_1=-\frac{1}{2},x_2=\sqrt{2},x_3=2{\log }_{3}4$  

Zatem:

$f\left(x_1\right)=f\left(-\frac{1}{2}\right)=3^{-\frac{1}{2}}={\left(\frac{1}{3}\right)}^{\frac{1}{2}}=\sqrt{\frac{1}{3}}=\frac{\sqrt{3}}{3}$  

$f\left(x_2\right)=f\left(\sqrt{2}\right)=3^{\sqrt{2}}$  

$f\left(x_3\right)=f\left(2{\log }_{3}4\right)=3^{2{\log }_{3}4}=3^{{\log }_3{4}^2}=4^2=16$  

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$c\text{)}f\left(x\right)={\left(0,2\right)}^x,x_1=-2,x_2={\log }_{5}3,x_3={\log }_5\frac{1}{12}$  

Zatem:

$f\left(x_1\right)=f\left(-2\right)={\left(0,2\right)}^{-2}={\left(\frac{10}{2}\right)}^2=5^2=25$  

$f\left(x_2\right)=f\left({\log }_{5}3\right)={0,2}^{{\log }_{5}3}={\left(\frac{1}{5}\right)}^{{\log }_{5}3}=$  

$=5^{-{\log }_{5}3}=5^{{\log }_5{3}^{-1}}=3^{-1}=\frac{1}{3}$  

$f\left(x_3\right)=f\left({\log }_5\frac{1}{12}\right)={0,2}^{{\log }_5\frac{1}{12}}={\left(\frac{1}{5}\right)}^{{\log }_5\frac{1}{12}}=5^{-{\log }_5\frac{1}{12}}=$  

$=5^{{\log }_5{\left(\frac{1}{12}\right)}^{-1}}={\left(\frac{1}{12}\right)}^{-1}=12$  

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$d\text{)}f\left(x\right)=4^{-x},x_1=-\frac{1}{2},x_2={\log }_{2}3,x_3={\log }_{4}7$  

Zatem:

$f\left(x_1\right)=f\left(-\frac{1}{2}\right)=4^{-\left(-\frac{1}{2}\right)}=4^{\frac{1}{2}}=\sqrt{4}=2$  

$f\left(x_2\right)=f\left({\log }_{2}3\right)=4^{-{\log }_{2}3}={\left(2^2\right)}^{-{\log }_{2}3}=$  

$=2^{-2{\log }_{2}3}=2^{{\log }_2{3}^{-2}}=3^{-2}=\frac{1}{9}$  

$f\left(x_3\right)=f\left({\log }_{4}7\right)=4^{-{\log }_{4}7}=4^{{\log }_4{7}^{-1}}=7^{-1}=\frac{1}{7}$