Find positive integers $a$ such that there are exactly distinct $2014$ positives integers $b$ satisfied $2 \le \dfrac{a}{b} \le 5$. 
Find positive integers $a$ such that there are exactly distinct $2014$ positives integers $b$ satisfied $2 \le \dfrac{a}{b} \le 5$.

This is another problem which I couldn't solve in the test my teacher gave us.
Here are my thoughts.
$2 \le \dfrac{a}{b} \le 5 \iff 2b \le a \le 5b$.
Let the possible $2014$ positive integers $b$ be $b_1 < b_2 < \cdots < b_{2013} <  b_{2014}$
$ \implies b_{2014} - b_1 \le 2013$.
We have that $2b_{2014} \le a \le 5b_1 \implies 2b_{2014} \le 5b_1$
$\implies \left\{ \begin{align} 2(b_1 + 2013) &\le 5b_1\\ 2b_{2014} &\le 5(b_{2014} - 2013) \end{align} \right.$ $\iff \left\{ \begin{align} 1342 &\le b_1\\ 3355 &\le b_{2014} \end{align} \right.$
$\implies b_1 = 1342, b_2 = 1343, \cdots, b_{2013} = 3354, b_{2014} = 3355$.
$\implies a = 2b_{2014} = 5b_1 = 6710$.
Is this the correct solution and is there any retouching needed?

Your solution isn't quite correct, as I discuss at the end. Instead, as trisct's comment indicates, it's easier to adjust your inequality to bound the $b$ values by multiples of $a$. In particular, with
$$2 \le \frac{a}{b} \le 5 \tag{1}\label{eq1B}$$
First, multiply the left & middle parts by $\frac{b}{2}$ to get
$$b \le \frac{a}{2} \tag{2}\label{eq2B}$$
Next, multiply the middle & right parts of \eqref{eq1B} by $\frac{b}{5}$ to get
$$\frac{a}{5} \le b \tag{3}\label{eq3B}$$
You can combine \eqref{eq2B} and \eqref{eq3B} into one set of inequalities of
$$\frac{a}{5} \le b \le \frac{a}{2} \tag{4}\label{eq4B}$$
For there to be exactly $2014$ values of $b$ (with all of them being consecutive) requires that the difference between the right & left sides of \eqref{eq4B} be at least $2013$, but less than $2015$. Thus, you get
$$2013 \le \frac{a}{2} - \frac{a}{5} = \frac{3a}{10} \lt 2015 \tag{5}\label{eq5B}$$
Multiplying all parts by $\frac{10}{3}$ gives
$$6710 \le a \lt 6716\;\frac{2}{3} \tag{6}\label{eq6B}$$
This gives up to $7$ possible values of $a$. However, it's possible some of them may not give the correct number of values of $b$, so you should check each one. To make it a bit simpler, let $a = 6710 + c$, with $0 \le c \le 6$. Then \eqref{eq4B} becomes
$$1342 + \frac{c}{5} \le b \le 3355 + \frac{c}{2} \tag{7}\label{eq7B}$$
For $c = 0$, you get $1342 \le b \le 3355$, so there's $2014$ values of $b$. However, for $c = 1$, you get $1342.2 \le b \le 3355.5$, so there's only $2013$ values of $b$ in this case. Continuing, you'll find that $c = 2,3$ work, while $c = 4,5,6$ don't. Thus, the final set of values of $a$ which work are $\{6710,6712,6713\}$.
Note this includes your solution of $6710$, and also includes $2$ others you didn't find. I see two issues with your written solution. First, you tried using $2b_{2014} \le a \le 5b_1$. Note the inequality is for each value of $b$, so you can't just use the largest on the left and the smallest on the right. Next, you have $2b_{2014} \le 5(b_{2014} - 2013) \iff b_{2014} \le 3355$. However, the correct result from this is $3355 \le b_{2014}$ instead.
