$a\text{)}\frac{8x}{x^2-8x}=\frac{8x}{x\cdot \left(x-8\right)}=\frac{8}{x-8}$    

---

$b\text{)}\frac{x^3+3x^2-2x-6}{x^2-2}=\frac{x^3-2x+3x^2-6}{x^2-2}=\frac{x\left(x^2-2\right)+3\left(x^2-2\right)}{x^2-2}=\frac{\left(x+3\right)\left(x^2-2\right)}{x^2-2}=x+3$  

---

$c\text{)}\frac{k-1}{k^2+k}\cdot \frac{2k}{k^2-1}=\frac{k-1}{k\cdot \left(k+1\right)}\cdot \frac{2k}{\left(k-1\right)\left(k+1\right)}=\frac{1}{k+1}\cdot \frac{2}{k+1}=\frac{2}{{\left(k+1\right)}^2}$  

---

$d\text{)}\frac{2x}{x+5}+\frac{10}{x+5}=\frac{2x+10}{x+5}=\frac{2\left(x+5\right)}{x+5}=2$  

---

$e\text{)}\frac{8-3x}{x+2}+3=\frac{8-3x}{x+2}+\frac{3\cdot \left(x+2\right)}{x+2}=\frac{8-3x}{x+2}+\frac{3x+6}{x+2}=\frac{8-3x+3x+6}{x+2}=\frac{14}{x+2}$  

---

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

---

$g\text{)}\frac{1}{x-4}-\frac{1}{x+4}=\frac{x+4-\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\frac{8}{x^2-16}$  

---

$h\text{)}\frac{x+8}{x}-\frac{x+5}{x-3}=\frac{\left(x-3\right)\left(x+8\right)-x\left(x+5\right)}{x\cdot \left(x-3\right)}=\frac{x^2+8x-3x-24-x^2-5x}{x\cdot \left(x-3\right)}=\frac{-24}{x^2-3x}$    

---

$i\text{)}\frac{a}{a-b}+\frac{b}{b-a}=\frac{a\left(b-a\right)+b\left(a-b\right)}{\left(a-b\right)\left(b-a\right)}=\frac{ab-a^2+ba-b^2}{\left(a-b\right)\left(b-a\right)}=\frac{-a^2+2ab-b^2}{\left(b-a\right)\left(a-b\right)}=\frac{-\left(a^2-2ab+b^2\right)}{-\left(a-b\right)\left(a-b\right)}=\frac{-{\left(a-b\right)}^2}{-{\left(a-b\right)}^2}=1$  

---

$j\text{)}\frac{x}{x^2-9}-\frac{1}{x+3}=\frac{x}{\left(x-3\right)\left(x+3\right)}-\frac{x-3}{\left(x+3\right)\left(x-3\right)}=\frac{x-\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{x-x+3}{\left(x+3\right)\left(x-3\right)}=\frac{3}{\left(x+3\right)\left(x-3\right)}=\frac{3}{x^2-9}$