Your argument is not completely correct, as you counted more than once some combination. This a way to fix it:
Number of ways to choose $3$ adjacent points : $32$
Number of ways to choose $3$ points such that $\textbf{only}$ two of them are adjacent : $32\cdot28$. Because there are $32$ couples of adjacent points and the third point must not be adjacent the other two.
Number of ways to choose $3$ points such that two of them are diametrically opposite but $\textbf{not adjacent}$: $16\cdot26$. Because the are $16$ total diameters and you can choose the third point in $26$ ways, as it must not be near the other two.
Answer: ${32 \choose 3 }-32\cdot28-32-16\cdot26=3616$
You appear to be off a bit: in your first case $3$ vertices are unavailable, not $2$.
I’ve numbered the vertices from $1$ through $2n$. For my first case I put a domino on vertices $1$ and $2$. Now I need to choose $k-1$ of the $2n-3$ vertices $3,4,\ldots,2n-1$, ensuring that no two chosen vertices are adjacent. This can be done in
$$\binom{(2n-3)-(k-2)}{k-1}=\binom{2n-1-k}{k-1}$$
ways.
For my second case I put a domino on vertices $2n$ and $1$; the analysis is the same, so we get another $\binom{2n-1-k}{k-1}$ arrangements.
Any other arrangement must avoid vertex $1$ entirely. In that case we need to choose $k$ of the $2n-2$ vertices $2,3,\ldots,2n-1$, ensuring that no two chosen vertices are adjacent. This can be done in
$$\binom{(2n-2)-(k-1)}k=\binom{2n-1-k}k$$
ways. I get a total of
$$2\binom{2n-1-k}{k-1}+\binom{2n-1-k}k=\binom{2n-1-k}{k-1}+\binom{2n-k}k$$
arrangements. I’ve checked this by hand with $n=4$ and $k=3$.
Best Answer
Yes, you are correct in thinking that the questions are connected. They all depend upon the method used in part (a).
(a) The chosen chairs occupy $k$ of the $n-k+1$ positions between and at the ends of the row of non-chosen chairs. The number of choices is therefore $$\begin{pmatrix}n-k+1\\k\\\end{pmatrix}.$$
(b) If chair 1 is not chosen, then the choice of $3$ chairs from the remaining $9$ is calculated as in (a). If chair 1 is chosen then, ignoring chair $1$ and the chairs adjacent to $1$, the choice of $2$ chairs from $7$ is also calculated as in part (a).
The number of choices is therefore $$\begin{pmatrix}7\\3\\\end{pmatrix}+\begin{pmatrix}6\\2\\\end{pmatrix}=50.$$
(c) As in part (b) we have $$\begin{pmatrix}n-k\\k\\\end{pmatrix}+\begin{pmatrix}n-k-1\\k-1\\\end{pmatrix}=\frac{n(n-k-1)!}{k!(n-2k)!}.$$