$\text{a)}$  

$\underset{x\to \frac{\pi }{6}}{\lim }\sin 2x=\sin \left(2\cdot \frac{\pi }{6}\right)=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$  

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$\text{b)}$  

$\underset{x\to \frac{\pi }{4}}{\lim }\left(\cos 2x+\sin x\right)=\cos \left(2\cdot \frac{\pi }{4}\right)+\sin \frac{\pi }{4}=\cos \frac{\pi }{2}+\sin \frac{\pi }{4}=0+\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$  

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$\text{c)}$  

$\underset{x\to \frac{\pi }{6}}{\lim }\frac{\mathrm{tg}x}{\sin x}=\frac{\mathrm{tg}\frac{\pi }{6}}{\sin \frac{\pi }{6}}=\frac{\frac{\sqrt{3}}{3}}{\frac{1}{2}}=\frac{\sqrt{3}}{6}$  

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$\text{d)}$  

$\underset{x\to \frac{\pi }{3}}{\lim }\left(\sin x-2\cos x\right)=\sin \frac{\pi }{3}-2\cos \frac{\pi }{3}=\frac{\sqrt{3}}{2}-2\cdot \frac{1}{2}=\frac{\sqrt{3}-2}{2}$