Statistics Projects for High School Students

Where should I use median instead of average?
Is there a general law, or rule of thumb, or rationale, when to use median and when average?
Although I know the difference and how they are computed, when I try to translate to simple English I would say in both cases that they both are a value that justly and fairly represents a big group of values of a certain category.
Examples:
Grades across different subjects for a single student.
Grades in one subject across students in one class.
Time to close a ticket per worker of a support desk.
Jail time given by a judge for a certain crime.
Lap time in a 10 laps run for a certain runner.
Monthly income per household in a given neighborhood.
So what should I use in each case above? And what is the general rule. And as a side question, are there other types of aggregate functions other than median and average that relate to this?

The problem is to find the slopes and $y$-intercepts of the following lines:
1) $x=-<!--\; -\; -->2$
and
2) $y=-<!--\; -\; -->1$.
I know that for the $y$-intercept we should set $x$ equal to zero and solve for $y$. For example, the y intercept for $y=2x+3$ is $(0,3)$ and slope is $2$. For the first equation, which is a line parallel to the y-axis and never crosses it, I think the y-intercept is undefined (or maybe zero) and also the slope is the same.
But for the second equation, the y-intercept is $-<!--\; -\; -->1$ (because it crosses $y$ at $-<!--\; -\; -->1$ point for all values of $x$). But I think that the slope is undefined. I think this because if we compare $y=-<!--\; -\; -->1$ to $y=mx+b$ (where $m$ is the slope), then we can't determine a simple slope.

Hypothesis Testing help
Really have no idea where to start :(
In an experiment comparing two weight-loss regimes A and B 20 test subjects were matched into 10 pairs so that within each pair the subjects were as similar as possible. Then A was randomly allocated to one of the subjects in each pair, and then B allocated to the other. The number of kilograms lost for each person is obtained and then the ordered A−B differences for each pair (in kg) are given below in the object d, together with some summary statistics:
sort(d) [1] -0.9 -0.3 0.2 0.4 0.6 1.2 1.4 3.3 3.5 4.3
mean(d) [1] 1.37
sd(d) [1] 1.755025
By specifying and checking (with a boxplot) an appropriate normality assumption perform a formal hypothesis test of H0:“regimes the same” against H1:“regimes not the same”.
Any help is appreciated!

This is equation of a curve $\sqrt{y}+\sqrt{x}=\sqrt{A}$
$A$ is a positive constant
$T$ is a tangent of the curve from any point on it
$B$ is the y-intercept of $T$
$C$ is the x-intercept of $T$
Prove that $B+C=A.$.

Given $f(x)=\frac{1}{4}(x+4{)}^2-<!--\; -\; -->2$
Find: vertex, $y$-intercept, $x$-intercepts (if any), axis of symmetry
What I have so far:
Vertex:$(-<!--\; -\; -->4,-<!--\; -\; -->2)$
$y$-intercept: $(0,2)$
$x$-intercept: $2$
Axis of symmetry $x=-<!--\; -\; -->4$

Find the $y$-intercept of the curve that passes through the point $(2,1)$ with the slope at $(x,y)$ of $\frac{-<!--\; -\; -->9}{y^2}$
$\frac{dy}{dx}=\frac{-<!--\; -\; -->9}{y^2}$
$\int <!--\; \int \; -->y^{2}dy=\int <!--\; \int \; -->-<!--\; -\; -->9dx$
$\frac{y^3}{3}=-<!--\; -\; -->9x+C_1$
$y^3=-<!--\; -\; -->27x+C$ $(C=3C_1)$
$y=(-<!--\; -\; -->27x+C{)}^{1/3}$
$1=(-<!--\; -\; -->27(2)+C{)}^{1/3}$
$C=54$
$0=(-<!--\; -\; -->27x+54{)}^{1/3}$
$y-<!--\; -\; -->intercept=(-<!--\; -\; -->2,0)$
Where is mistake?

What is the lower quartile of the set of data?
I came across this problem asking the lower quartile of the ungrouped data. My answer is 3, but other references say it should be 2.5. Here's the data:
1, 1, 2, 2, 3, 3, 4, 4, 6, 7, 8, 10, 11, 14, 15, 20, 22
What do you think?

The gradient is $m=-<!--\; -\; -->4/3$, and the point given is $(3/2,3)$
Let $c$ be the $y$-intercept
$y=mx+c$
$y=(-<!--\; -\; -->4/3)x+c$
If I substitute $x$ and $y$ from the point given, $3=(-<!--\; -\; -->4/3)(3/2)+c$ and look for $c$, I get $c=5$
But according to my answer sheet, $c=33/8$ by using $y-<!--\; -\; -->y_1=m(x-<!--\; -\; -->x_1)$
Why are the two methods giving different answers?

Calculating the position of the median
I know this is a simple question, but I cannot find a straight answer anywhere.
When calculating the medium of listed data, the formula is (n+1)/2. My Statistics teacher said for grouped data the position of the medium is n/2. However, this seems contradictory as for discrete grouped data, the data could be written in a list if the original values were known. Therefore two different values for the median are found.
I know this question is probably going to get flagged as it has already been asked, however, it has never been answered. I find it frustrating that such a fundamental concept in Statistics, what is supposed to be precise and never subjective has a wishy-washy answer.
Rant over, I think the position should be (n+1)/2 for discrete data and n/2 for continuous data. Moreover, this then leads into the obvious question of how we should calculate quartiles as well.
It would be nice if everyone could agree on a certain method for calculating the medium as all my textbooks are saying different things.
Someone please dispell the confusion and this statistical mess.

An equation to the tangent line of $h(x)=\tan <!--\; \; -->(x)+\cos <!--\; \; -->(x)$ is given by $y=mx+b$ where $m$ is the slope and $b$ is the $y$-intercept. If $x=\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$ find $m$.

calculating upper and lower quartiles
The number of goals scored by a football team during a months worth of matches is recorderd.
They scored $0$ goals $4$ times, $1$ goal $6$ six times and $2$ goals $3$ times. Calculate the lower and upper quartiles.
Would it simply be this?: lower quartile is the $3.5$th value which is zero, and upper quartile is the $10.5$th value which is $2$?

Two questions involving quadratics. Next to a diagram of a parabola with a ma point with the y intercept of $(0,p)$ and roots $(-<!--\; -\; -->1,0)$ and $(p,0)$ it says:3a) Show that the equation of the parabola is $y=p+(p-<!--\; -\; -->1)x-<!--\; -\; -->x^2$
3b) For what value of $p$ will the line $y=x+p$ be a tangent to this curve.

Need to find the $y$-intercept of the tangent to $y=\frac{18}{x^2+2}$ at point $(1,6)$.
I keep getting $0$, but I don't think that's right.