Which fraction $\dfrac{p}{q}$, where $p,q$ are positive integers less than $100$, is closest to $\sqrt{2}$? 
Which fraction $\dfrac{p}{q}$, where $p,q$ are positive integers less than $100$, is closest to $\sqrt{2}$? Find all digits after the decimal point in the decimal representation of this fraction that coincided with digits in the decimal representation of $\sqrt{2}$ (without using any tables).

Firstly, what do they mean by "without using any tables"? Do they mean without just remembering what $\sqrt{2}$ is? Also, in the solution below how do they get the bound $7 \cdot 10^{-5} < \dfrac{p}{q}-\sqrt{2} < 8 \cdot 10^{-5}$?
Book's solution: 

 A: There is a more clear way to solve this by using the simple continued fraction for $x=\sqrt{2}$:
$$x^2=2$$
$$x^2-1=1$$
$$(x-1)(x+1)=1$$
$$x-1=\frac{1}{1+x}$$
$$x=1+\frac{1}{1+x}=1+\cfrac{1}{1+1+\cfrac{1}{1+x}}=1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{1+x}}}=\dots$$
Finally, we obtain the simple continued fraction for $\sqrt{2}$:
$$\sqrt{2}=1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cdots}}}}$$
Now, all the best rational approximations of $\sqrt{2}$ (with the smallest possible denominator) can be obtained from this continued fraction by truncating it.
This way we get the following sequence of approximants:
$$\frac{p_k}{q_k}=\{1, \frac{3}{2} , \frac{7}{5}, \frac{17}{12}, \frac{41}{29},  \frac{\bf 99}{\bf 70} , \frac{239}{169},\dots \}$$
As you can see, the answer is really $\frac{99}{70}$, because the better approximation has $p,q>100$.

This way is a little long, and you need to take for granted the fact that a simple continued fraction really gives the best rational approximants.
But I agree that the 'solution' in your textbook is not very clear.
See also https://en.wikipedia.org/wiki/Square_root_of_2 for reference.
