a)

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

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

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

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

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

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

$f\left(2\right)=\frac{\sqrt{6}}{2}$  

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b)

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

$\frac{16}{81}={\left(\frac{2}{3}\right)}^{x-1}$  

${\left(\frac{2}{3}\right)}^4={\left(\frac{2}{3}\right)}^{x-1}$  

$4=x-1\mid +1$  

$5=x$  

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c)

$f\left(x\right)>2\frac{1}{4}$  

${\left(\frac{2}{3}\right)}^{x-1}>2\frac{1}{4}$  

${\left(\frac{2}{3}\right)}^{x-1}>\frac{9}{4}$  

${\left(\frac{2}{3}\right)}^{x-1}>{\left(\frac{2}{3}\right)}^{-2}$  

Zauważmy, że funkcja  $y={\left(\frac{2}{3}\right)}^x$  jest funkcją malejącą, zatem należy zmienić znak nierówności.

$x-1<-2\mid +1$  

$x<-1$  

$x\in \left(-\infty ,-1\right)$  

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d)

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

$g\left(x\right)=f\left(-x\right)-3$  

$g\left(x\right)={\left(\frac{2}{3}\right)}^{-x-1}-3$  

$g\left(x\right)={\left(\frac{3}{2}\right)}^{x+1}-3$  

Wykres funkcji g(x):

$x$	$-1$	$0$	$1$
$g\left(x\right)$	${\left(\frac{3}{2}\right)}^{-1+1}-3=$   $={\left(\frac{3}{2}\right)}^0-3=$   $=1-3=-2$	${\left(\frac{3}{2}\right)}^{0+1}-3=$   $=\frac{3}{2}-3=-1\frac{1}{2}$	${\left(\frac{3}{2}\right)}^{1+1}-3=$   $={\left(\frac{3}{2}\right)}^2-3=$   $=\frac{9}{4}-3=$   $=\frac{9}{4}-\frac{12}{4}=-\frac{3}{4}$