$textbf{(a)}$      For the first $5 text{s}$, the float moves at a constant speed of $2 frac{text{m}}{text{s}}$ in a straight line. This corresponds to the first straight line slanting upwards in the position-time graph.

After that, the float comes to a $5 text{s}$ stop. Since the position doesn’t change with time, this segment is represented by a straight horizontal line.

Finally, the float starts moving at $1 frac{text{m}}{text{s}}$ in the initial direction, which corresponds to the second straight line slanting upwards.

$textbf{(b)}$      To find the position of the float at a given time, find that value on the $x$ axis and trace from that point upward until you reach the straight line. From that point, trace sideways until you reach the $y$ axis.

At $t=2 text{s}$, the position of the float is

$$
begin{align*}
boxed{x_{2text{s}}=4 text{m}}\
end{align*}
$$

and at $t=11 text{s}$, the position of the float is

$$
begin{align*}
boxed{x_{11text{s}}=11 text{m}}\
end{align*}
$$

$$
begin{align*}
textbf{(a)} quad & text{Click to see the graph.}\
\
textbf{(b)} quad &boxed{x_{2text{s}}=4 text{m}}\
&boxed{x_{11text{s}}=11 text{m}}\
end{align*}
$$