Given $n$ people, where $n$ is even, you can choose the first pair of people ${n \choose 2}$ ways, where ${n \choose 2}=\frac{n!}{2!(n-2)!}$. The next pair can be chosen in ${n-2\choose 2}$ ways, etc... The end result will be $\frac{n}{2}$ pairs which can be arranged in $\left(\frac{n}{2}\right)!$ ways. So there are $$\frac{{n \choose 2}{n-2 \choose 2}\dots{2 \choose 2}}{\left(\frac{n}{2}\right)!}=\frac{n!}{\left(\frac{n}{2}\right)!\,2^{\left(\frac{n}{2}\right)}}$$ ways to arrange all $n$ people into sets of pairs.
So for 8 people there are $\frac{8!}{4!\,2^{4}}=105$ possible sets of pairs.
Now the question remains, how many sets are there that do not contain any pairs from the original set?
Let $$f(x)=\frac{x!}{\left(\frac{x}{2}\right)!\,2^{\frac{x}{2}}}$$
If $S=\left\{p_{1}, p_{2}, ..., p_{\frac{n}{2}}\right\}$ is our original set of pairs and $P_{k}$ is the set of all sets containing $p_{k}$, where $1\le k\le\frac{n}{2}$ then by the inclusion-exclusion principal the number of sets not containing any of the original pair is:
$$f(n)-\left(|P_{1}|+|P_{2}|+\dots+|P_{\frac{n}{2}}|\right)+\left(|P_{1}\cap P_{2}|+(|P_{1}\cap P_{3}|+\dots\right)-\dots$$
But for any $P_{k}$; $|P_{k}|=f(n-2)$ therefore, $|P_{1}|=|P_{2}|=\dots=|P_{\frac{n}{2}}|$and in general, given any $k$ where $1\le k\le \frac{n}{2}$, then $|P_{1}\cap P_{2}\cap\dots\cap P_{k}|=f(n-2k)-\dots$
So the number of sets not containing any of the original pair is:
$$f(n)-\left(f(n-2)+f(n-2)+\dots\right)+\left(f(n-4)+f(n-4)+\dots\right)-\dots$$
which equals:
$$f(n)-{\frac{n}{2}\choose 1}f(n-2)+{\frac{n}{2}\choose 2}f(n-4)-\dots$$
So in the case where $n=8$
$$f(8)-4f(6)+6f(4)-4f(2)+1f(0)=105-60+18-4+1=60$$
In graph theory terms, you would like to partition a complete graph into matchings ($1$-regular graphs). There is a well-known algorithm for doing that. Your students would be the vertices of this graph, and a pair of students working together would be an edge of this graph.
First, label your $n$ students (where $n$ is even) as follows: $\infty,0,1,2,\dots,n-2$. Think of the finite (non-$\infty$) labels as remainders modulo $n-1$. Now, for each $i=0,1,2,\dots,n-2$, pair your students as follows: $(\infty,i)$ and $(i+k,i-k)$ for $k=1,2,\dots,\frac{n}{2}-1$. In other words, except for the special pair with $\infty$, the sum of the two labels in a pair should be $2i$ modulo $n-1$.
For example, let $n=8$, then $n-1=7$ (so all labels are remainders modulo $7$), and your vertex labels are $\infty,0,1,2,3,4,5,6$. The matchings you need are as follows:
$$
\begin{matrix}
i=0: & (\infty,0), (1,6), (2,5), (3,4)\\
i=1: & (\infty,1), (2,0), (3,6), (4,5)\\
i=2: & (\infty,2), (3,1), (4,0), (5,6)\\
i=3: & (\infty,3), (4,2), (5,1), (6,0)\\
i=4: & (\infty,4), (5,3), (6,2), (0,1)\\
i=5: & (\infty,5), (6,4), (0,3), (1,2)\\
i=6: & (\infty,6), (0,5), (1,4), (2,3)
\end{matrix}
$$
To see that this as a picture, imagine the labels $0,1,2,\dots,n-2$ as spaced evenly around a circle while $\infty$ is placed in the center. Then $(\infty,i)$ is a "minute hand" pointing at $i$, while the rest of the pairs are all edges perpendicular to it.
Best Answer
For the first part, you have already given a nice intuitive explanation.
For an intuitive explanation of the second formula, in the $(2n)!$ permutations of the items which we can pair serially, neither the order of the pairs nor the order within the pairs matter, thus the division by $n!2^n$,
I much prefer, though, to think of the ways for sequential choices being $[(2n-1)(2n-3)(2n-5)...1 = (2n-1)!!$,