a) Mamy

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

Stąd

$f\left(x_1\right)={\left(\frac{4}{3}\right)}^{-2}={\left(\frac{3}{4}\right)}^2=\frac{9}{16}$  

$f\left(x_2\right)={\left(\frac{4}{3}\right)}^0=1$  

$f\left(x_3\right)={\left(\frac{4}{3}\right)}^2=\frac{16}{9}$  

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b) Mamy

$f\left(x\right)={\left(1\frac{7}{9}\right)}^x={\left(\frac{16}{9}\right)}^x$  

Stąd

$f\left(x_1\right)={\left(\frac{16}{9}\right)}^{-2}={\left(\frac{9}{16}\right)}^2=\frac{81}{256}$  

$f\left(x_2\right)={\left(\frac{16}{9}\right)}^{-\frac{1}{2}}={\left(\frac{9}{16}\right)}^{\frac{1}{2}}=\frac{3}{4}$  

$f\left(x_3\right)={\left(\frac{16}{9}\right)}^{\frac{1}{4}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$  

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c) Mamy

$f\left(x\right)=4^x$  

Skorzystamy ze wzorów

$a^{{\log }_{a}b}=b$  

$n\cdot {\log }_{a}b={\log }_a{b}^n$  

Stąd

$f\left(x_1\right)=4^{{\log }_{4}5}=5$  

$f\left(x_2\right)=4^{{\log }_{2}7}=2^{2{\log }_{2}7}=2^{{\log }_2{7}^2}=7^2=49$  

$f\left(x_3\right)=4^{{\log }_{2}0,5}=2^{2{\log }_{2}0,5}=2^{{\log }_2{\left(\frac{1}{2}\right)}^2}=\frac{1}{2^2}=\frac{1}{4}$  

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d) Mamy

$f\left(x\right)={81}^{-x}=3^{-4x}$  

Stąd

$f\left(x_1\right)=3^{-4\cdot \frac{1}{2}}=3^{-2}=\frac{1}{9}$  

$f\left(x_2\right)=3^{-4\cdot \frac{1}{3}}=\frac{1}{3^{1\frac{1}{3}}}=\frac{1}{3\sqrt[3]{3}}$  

$f\left(x_3\right)=3^{-4\cdot \frac{1}{4}}=3^{-1}=\frac{1}{3}$