$\text{a)}$  

$\underset{x\to 3}{\lim }\frac{x^3-x^2-5x-3}{x^3+x-3}=\frac{3^3-3^2-5\cdot 3-3}{3^3+3-3}=\frac{0}{27}=0$      

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$\text{b)}$  

$\underset{x\to 2}{\lim }\frac{x^3-8}{x^2-x-2}=\underset{x\to 2}{\lim }\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+1\right)\left(x-2\right)}=\underset{x\to 2}{\lim }\frac{x^2+2x+4}{x+1}=$  

$=\frac{2^2+2\cdot 2+4}{2+1}=\frac{12}{3}=4$  

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$\text{c)}$  

$\underset{x\to 0}{\lim }\frac{\sin 2x}{\sin x}=\underset{x\to 0}{\lim }\frac{2\sin x\cos x}{\sin x}=\underset{x\to 0}{\lim }2\cos x=2\cdot \cos 0=2\cdot 1=2$  

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$\text{d)}$  

$\underset{x\to -1}{\lim }\frac{x+1}{\sqrt{x+2}+x}=\underset{x\to -1}{\lim }\frac{\left(x+1\right)\left(\sqrt{x+2}-x\right)}{\left(\sqrt{x+2}+x\right)\left(\sqrt{x+2}-x\right)}=$  

$=\underset{x\to -1}{\lim }\frac{\left(x+1\right)\left(\sqrt{x+2}-x\right)}{{\sqrt{x+2}}^2-x^2}=\underset{x\to -1}{\lim }\frac{\left(x+1\right)\left(\sqrt{x+2}-x\right)}{\left|x+2\right|-x^2}=$   

$\underset{x\to -1}{\lim }\frac{\left(x+1\right)\left(\sqrt{x+2}-x\right)}{-x^2+x+2}=\underset{x\to -1}{\lim }\frac{\left(x+1\right)\left(\sqrt{x+2}-x\right)}{-\left(x+1\right)\left(x-2\right)}=$  

$=\underset{x\to -1}{\lim }\frac{\sqrt{x+2}-x}{-\left(x-2\right)}=\frac{\sqrt{-1+2}-\left(-1\right)}{-\left(-1-2\right)}=\frac{1+1}{3}=\frac{2}{3}$