a)

$a_n=2n+1,b_n=3n,n\in {\mathbb{N}}_{+}$

$a_{10}=2\cdot 10+1=20+1=21$

$b_8=3\cdot 8=24$

Zatem:

$a_{10}-b_8=21-24=-3$     

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b) 

$a_n=\frac{8-n}{4},b_n={\left(-1\right)}^{n+1}\cdot \frac{n-1}{2},n\in {\mathbb{N}}_{+}$

$a_{10}=\frac{8-10}{4}=-\frac{2}{4}=-\frac{1}{2}$

$b_8={\left(-1\right)}^{8+1}\cdot \frac{8-1}{2}={\left(-1\right)}^9\cdot \frac{7}{2}=-1\cdot \frac{7}{2}=-\frac{7}{2}$

Zatem:

$a_{10}-b_8=-\frac{1}{2}-\left(-\frac{7}{2}\right)=-\frac{1}{2}+\frac{7}{2}=\frac{6}{2}=3$     

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c)

$a_n=\frac{n^2+5}{n},b_n=\frac{5n-8}{n^2},n\in {\mathbb{N}}_{+}$  

$a_{10}=\frac{{10}^2+5}{10}=\frac{100+5}{10}=\frac{105}{10}=\frac{21}{2}$

$b_8=\frac{5\cdot 8-8}{8^2}=\frac{40-8}{64}=\frac{32}{64}=\frac{1}{2}$

Zatem:

$a_{10}-b_8=\frac{21}{2}-\frac{1}{2}=\frac{20}{2}=10$     

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d)

$a_n=\sqrt[3]{2-n},b_n=n^{\frac{2}{3}},n\in {\mathbb{N}}_{+}$

$a_{10}=\sqrt[3]{2-10}=\sqrt[3]{-8}=-2$

$b_8=8^{\frac{2}{3}}={\left(\sqrt[3]{8}\right)}^2=2^2=4$

Zatem:

$a_{10}-b_8=-2-4=-6$