To find $q$, notice that the parabola has only one root, and that this occurs exactly halfway along the width of the riverbed.

To find $p$, notice that the point $(10,4)$ lies on the parabola.

For (b), the diagram should look like this

Note that $P$ and $Q$ have a $y$ coordinate of $1.96$

For (c), the diagram looks like this

You can use the symmetry to find the $x$ coordinates of the shoreline (where the surface of the water meets the riverbank). Once you have the $x$ coordinates, find the $y$ coordinates using the equation of the curve.

1. The parabola has only one root, when $x=5$ (we determine this by symmetry). The equation $p(x-q{)}^2=0$ is only solved by $x=q$, and so we conclude that $q=5$ and $y=p(x-5{)}^2$.

   When $x=0,y=4$ and so

   $$\begin{array}{rl}4& =p(0-5{)}^2\\ 25p& =4\\ p& =\frac{4}{25}\end{array}$$

2. The situation described looks like this:

   

   When $y=1.96$, we have

   $$\begin{array}{rl}1.96& =\frac{4}{25}(x-5{)}^2\\ 49& =4(x-5{)}^2\\ x-5& =\pm \sqrt{\frac{49}{4}}\\ x& =5\pm \frac{7}{2}\end{array}$$

   The $x$ coordaintes of $P$ and $Q$ are $\frac{3}{2}$ and $\frac{17}{2}$ so the distance between them is $7$.

3. We're now dealing with this situation

   

   By considering the symmetry, we see that the $x$ coordinates where the water meets the riverbed are $5\pm 3$. Substituting either of these into the equation of the parabola will give us $h$, for example

   $$\begin{array}{rl}h& =\frac{4}{25}(8-5{)}^2\\ & =\frac{36}{25}\end{array}$$