Both of your solutions to the first problem are correct. However, your general formula is not. Notice that in the first problem, $n = 8$, $m = 5$, and $k = 2$, so your formula gives
$$2!P(8 - 2 + 1, 8 - 5) = 2!P(7, 3) = 2! \cdot 7 \cdot 6 \cdot 5 = 2 \cdot 210 = 420$$
Let's see what went wrong.
We wish to seat $m$ people, $k$ of whom must sit consecutively, in $n$ seats. Since the block takes up $k$ of the $n$ places, it must begin in one of the first $n - (k - 1) = n - k + 1$ positions. Once the block has been placed, there are $n - k$ seats left for the remaining $m - k$ people. They can be arranged in those seats in $P(n - k, m - k)$ ways. The people within the block can be arranged in $k!$ ways, which gives us the formula
$$(n - k + 1)P(n - k, m - k)k!$$
As a sanity check, let's try our formula when $n = 8$, $k = 2$, and $m = 5$. It gives
$$(8 - 2 + 1)P(8 - 2, 5 - 2)2! = 7 \cdot P(6, 3) \cdot 2! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 2 = 1680$$
which agrees with the answer you obtained in your example.
Observe that
\begin{align*}
(n - k + 1)P(n - k, m - k) & = (n - k + 1) \cdot \frac{(n - k)!}{[(n - k) - (m - k)]!}\\
& = \frac{(n - k + 1)(n - k)!}{(n - m)!}\\
& = \frac{(n - k + 1)!}{(n - m)!}\\
& = \frac{(n - k + 1)!}{[(n - k + 1) - (m - k + 1)]!}\\
& = P(n - k + 1, m - k + 1)
\end{align*}
so we could write our formula in the form
$$P(n - k + 1, m - k + 1)k!$$
As a sanity check, note that if $n = 8$, $k = 2$, and $m = 5$, then
$$P(8 - 2 + 1, 5 - 2 + 1)2! = P(7, 4)2! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 2 = 1680$$
Best Answer
It's more simple than you are making it.
You can choose any two adjacent seats (no gap). There are $26$ such pairs.
You can choose any three adjacent seats (one gap). There are $25$ such triples.
So there are $51$ total positionings, but they can be flipped, giving a true total of $102$.