Okres podstawowy funkcji f jest równy T=9, więc:

$\text{a)}f\left(10\right)=f\left(9+1\right)=f\left(1\right)=g\left(1\right)=4$  

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$\text{b)}f\left(50\right)=f\left(54-4\right)=f\left(6\cdot 9-4\right)=f\left(-4\right)=g\left(-4\right)=2$  

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$\text{c)}f\left(-91\right)=f\left(-90-1\right)=f\left(-10\cdot 9-1\right)=f\left(-1\right)=g\left(-1\right)=4$  

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$\text{d)}f\left(544\frac{1}{2}\right)=f\left(540+4\frac{1}{2}\right)=f\left(60\cdot 9+4\frac{1}{2}\right)=f\left(4\frac{1}{2}\right)=g\left(4\frac{1}{2}\right)=0$  

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$\text{e)}\frac{f\left(20\right)+f\left(30\right)}{f\left(40\right)-f\left(50\right)}=\frac{f\left(18+2\right)+f\left(27+3\right)}{f\left(36+4\right)-f\left(54-4\right)}=\frac{f\left(2\cdot 9+2\right)+f\left(3\cdot 9+3\right)}{f\left(4\cdot 9+4\right)-f\left(6\cdot 9-4\right)}=\frac{f\left(2\right)+f\left(3\right)}{f\left(4\right)-f\left(-4\right)}=\frac{g\left(2\right)+g\left(3\right)}{g\left(4\right)-g\left(-4\right)}=\frac{2+0}{-2-2}=\frac{2}{-4}=-\frac{1}{2}$  

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$\text{f)}\left[f\left(-80\right)+f\left(80\right)\right]\cdot f\left(800\right)=\left[f\left(-81+1\right)+f\left(81-1\right)\right]\cdot f\left(801-1\right)=\left[f\left(-9\cdot 9+1\right)+f\left(9\cdot 9-1\right)\right]\cdot f\left(89\cdot 9-1\right)=\left[f\left(1\right)+f\left(-1\right)\right]\cdot f\left(-1\right)=\left[g\left(1\right)+g\left(-1\right)\right]\cdot g\left(-1\right)=\left(4+4\right)\cdot 4=8\cdot 4=32$