What is the smallest possible value of $q$ such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$?

The way to find the smallest denominator "from scratch" is with continued fractions.

Begin by rendering the proposed bounds thusly:

$\dfrac{7}{10}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{3}}}}$

$\dfrac{11}{15}=\dfrac{1 }{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{1+\dfrac{1}{3}}}}}$

The upper layers of the continued fractions are identical but they eventually become different when we get down to the layers in blue. We may now replace those entries with the smallest whole number lying between, thus

$1+\dfrac{1}{3}<2<3$

So the smallest denominator fraction meeting the betweenness criterion will be

$\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{2}}}}=\dfrac{5}{\color{blue}{7}}$

If $7$ worked we would have

\begin{align} &\frac{21}{30}<\frac{p}{7}< \frac{22}{30}\\ \iff&\frac{21*7}{30*7}<\frac{30p}{30*7}< \frac{22*7}{30*7} \end{align} So $7$ is a solution if and only if there is a multiple of $30$ between $21*7$ and $22*7$. Since $21*7=147$ and $22*7=154$ we have a multiple of $30,$ namely $150$. So $p=\frac{5}{7}$ is between those numbers.

To prove that this is the smallest just notice that every number smaller than $7$ other than $4$ is a divisor of $30$ so their fractions can be written as $\frac{p}{30}.$ And obviously it is impossible to have

\begin{align*} \frac{21}{30}<\frac{p}{30}<\frac{22}{30} \end{align*}

As for $4$ if $4$ was possible we would have

\begin{align} &\frac{21}{30}<\frac{p}{4}<\frac{22}{30}\\ \iff & \frac{21*2}{30*2}<\frac{p*15}{4*15}<\frac{22*2}{30*2} \end{align} and there are no multiples of $15$ between $42$ and $44$.