Let me first say that, while this may appear to be a long and complicated question, if you break it into parts it is not.
Let's see what information we can learn from the problem.
Assume the depth of the water is a sinusoidal function of time
OK, so that gives us a huge leap. What are some examples of sinusoidal functions? Certainly $f(t)=\sin(t)$ qualifies, but this is but the simplest one. We also have: $f(t)=23\sin(t)\ ,\qquad f(t)=\sin(68t)\ , \qquad f(t)=\sin(t+4)\ ,\qquad f(t)=\sin(t)+95$ or any combination of the above, using any constants in place of the explicit numbers. How can we write this most generally? Well, let $A,\ B,\ ,C$ and $D$ be constants. Then, in general, a sinusoidal function will be given by: $f(t)=A\sin\left(B(t+C)\right)+D\ .$ We've exhausted the information we can glean from that statement, let's move on.
High Tide: 11 ft deep
Low Tide: 7 ft deep
This means that the largest value of $f(t)$ is 11, and the smallest is 7. Since sine oscillates between 1 and -1, we know that this information will not help us find $B$ or $C$. On the other hand, since we know how sine behaves, we know that the smallest value of $f(t)$ will be $A(-1)+D$ (when the sine function is at its lowest point) and the largest value of $f(t)$ will be $A(1)+D$, (when the sine function is at its highest point). We now need to solve
$A+D=11\ ,$ $D-A=7\ .$
You can do this a number of ways; for instance, add the first equation to the second and you will find that $2D=18$, so we have $D=9$. Substitute this value of $D$ into either equation to arrive at $A=2$.
It is important to tie this result to the graphical idea of what we are doing. The value of $A$ is called the amplitude, and it will "stretch" the sine function vertically, as shown here:



The value of $D$ will move the sine function vertically:



So we now have pinned down $A$ and $D$, and we have $f(t)=2\sin(B(t+C)+9\ .$ We are given that
The period is half a lunar day (12hr, 24min)
It will do us no good in mixed form like that, so we'll convert it to 744 minutes. We had not previously mentioned what value $t$ describes, but I think at this point it makes sense to remark that it will be the number of minutes after the (literal) start of Labor Day.
Now, surely you are aware that the period of the sine function is $2\pi$, a beautiful number mathematically, but fairly ugly in this context. We need the period to be 744 minutes, and one way to think about this is that we want to stretch the sine function horizontally so that the value that used to be at $2\pi$ is now at 744. Let's look at just $g(t)=\sin(Bt)$. If we wanted the value of $g(744)$ to be the same as the value of $\sin(2\pi)$, we can just try to figure out how to make $B(744)=2\pi$. But this is easy, we can just solve for $B=\frac {2\pi} {744}$. The same thing happens with our more complicated function (that's why I used $B$ again), so we now have $f(t)=2\sin\left(\frac {2\pi} {744} (t+C)\right) + 9\ .$
There's only one thing left to address, our function needs to be oriented horizontally. This is called the "offset". Conceptually in our problem, you can think of the offset as describing the position of the waves at a specific time. Right now, we are starting with the waves at 9 feet, since $f(0)=9$ (as you can verify), but nothing in the problem tells us what height the waves started at at the beginning of the day. We do have the height of the waves at 7:12am, though, which is 432 minutes after the beginning of the day. We are told
At 7:12am, [...] the water is 11 feet deep.
So we need $f(432)=11$. This is a little complicated to calculate, but you can think about it like this: the high tide needs to happen at $t=432$. The sine function will describe a "high tide" 1/4 of the way through its period (remember that $\sin(\pi/4)=1$?), and our function $f$ has a period of 744, so (without adjusting C) it will have a "high tide" at 186, which is not what we want. We want $432+C=186$, which implies that $C=-246$. We now have the whole function:
$f(t)=2\sin\left(\frac {2\pi} {744} (t-246)\right) + 9$
Again, the geometric notion of what $C$ does is important - as you probably know, changing $C$ is moving the graph horizontally to the left or the right. If $C$ is negative, the graph moves to the right, so we have shifted the graph to the right in an effort to make the high-tide line up with 7:12am.



Now that you have the function, I leave it to you to answer the rest of the questions.