Hello Hey,
For these problems, we may rely on the following result:
$$\lim_{x\to0}\frac{\sin(x)}{x}=1$$
This is the answer to the first problem.
For the first two problems, let's develop a general formula to handle limits of the type:
$$\lim_{x\to0}\frac{\sin(ax)}{bx}$$
where $a$ and $b$ are non-zero real constants.
If we multiply the expression by $$1=\frac{a/b}{a/b}$$ and use the limit property:
$$\lim_{x\to c}k\cdot f(x)=k\cdot\lim_{x\to c}f(x)$$ where $k$ is a real constant
Then our limit becomes:
$$\frac{a}{b}\lim_{x\to0}\frac{\sin(ax)}{ax}$$
Now, using the substitution:
$$u=ax$$
and observing:
$$x\to0$$ implies $$u\to0$$
we may write:
$$\frac{a}{b}\lim_{u\to0}\frac{\sin(u)}{u}=\frac{a}{b}$$
Hence, we have found:
$$\lim_{x\to0}\frac{\sin(ax)}{bx}=\frac{a}{b}$$
And so the second limit is:
$$\lim_{x\to0}\frac{\sin(2x)}{6x}=\frac{2}{6}=\frac{1}{3}$$
For the third problem, let's consider the following limit:
$$\lim_{x\to0}\frac{\sin(ax)}{\sin(bx)}$$
We may then write:
$$\frac{\sin(ax)}{\sin(bx)}=\frac{a}{b}\frac{\frac{\sin(ax)}{ax}}{\frac{\sin(bx)}{bx}}$$
and making use of the limit property:
$$\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{\lim_{x\to c}f(x)}{\lim_{x\to c}g(x)}$$
we may write:
$$\lim_{x\to0}\frac{\sin(ax)}{\sin(bx)}=\frac{a}{b} \cdot\frac{\lim\limits_{x\to0}\frac{\sin(ax)}{ax}}{\lim\limits_{x\to0}\frac{\sin(bx)}{bx}}$$
And the using the substitutions $$u=ax,\,v=bx$$ we have:
$$\lim_{x\to0}\frac{\sin(ax)}{\sin(bx)}=\frac{a}{b} \cdot\frac{\lim\limits_{u\to0}\frac{\sin(u)}{u}}{ \lim\limits_{v\to0}\frac{\sin(v)}{v}}= \frac{a}{b}\cdot\frac{1}{1}$$
And so we may write:
$$\lim_{x\to0}\frac{\sin(ax)}{\sin(bx)}=\frac{a}{b}$$
And so the third limit is:
$$\lim_{x\to0}\frac{\sin(7x)}{\sin(5x)}=\frac{7}{5}$$