Your way1 does not count arrangements with many chairs next to one another, for example it doesn't count the arrangement |P|CCCC|P|C|P|C|P|, instead only allowing a single "extra" chair to be placed in each empty spot. Use stars and bars instead for how to distribute the extra chairs.
Approaching the same, we have $\underline{~}PC\underline{~}PC\underline{~}PC\underline{~}P\underline{~}$ and we wish to distribute three extra chairs to these five empty spaces (with possibly multiple going to the same empty spot).
By stars and bars we have $\binom{5+3-1}{5-1}=\binom{7}{4}$ number of ways to do so. Multiplying by the number of ways to arrange the persons themselves among the seats labeled $P$ we have a total of $4!\binom{7}{4}=~^7P_4=840$ arrangements, same as the other answer.
Method 1: We begin by arranging three blue balls and three red balls in a row. There are $\binom{6}{3}$ such arrangements since we must choose which three of the six spaces will be occupied by the blue balls. This creates seven spaces in which we can place three green balls, five spaces between successive balls and two at the ends of the row. For instance,
$$\square \color{red}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{red}{\bullet} \square \color{blue}{\bullet} \square \color{red}{\bullet} \square$$
We now wish to insert three green balls so that no two of them are adjacent. To do so, we must choose three of these seven spaces, which can be done in $\binom{7}{3}$ ways. One such arrangement is
balls and two at the ends of the row. For instance,
$$\color{green}{\bullet} \color{red}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{red}{\bullet} \color{blue}{\bullet} \color{red}{\bullet}$$
The green balls represent the positions that can be occupied by Alvin, John, and Albert; the blue balls represent the positions of the other three people; the red balls represent the positions of the three unoccupied seats. We can arrange Alvin, John, and Albert in the places occupied by green balls in $3!$ ways. We can arrange the remaining three people in the positions occupied by the blue balls in $3!$ ways.
Consequently, the number of permissible seating arrangements is
$$\binom{6}{3}\binom{7}{3}3!3!$$
Method 2: We use the Inclusion-Exclusion Principle.
There are $\binom{9}{3}$ ways of choosing the locations of the empty seats and $6!$ ways of arranging the six people in the remaining seats. From these seating arrangements, we must exclude those arrangements in which the three students who do not wish to sit in adjacent seats sit in adjacent seats.
If two of Alvin, John, and Albert sit together, we have eight objects to arrange, the pair of seats in which we will place those students, the three empty seats, and the other four students. There are $\binom{8}{3}$ ways to choose three of the eight positions for the empty seats, $\binom{3}{2}$ ways to choose two of the three students to sit together, $5!$ ways to arrange the remaining objects, and $2!$ ways to arrange the pair of chosen students in the designated pair of seats.
$$\binom{8}{3}\binom{3}{2}5!2!$$
Subtracting this from the total removes those cases in which all three students sit together twice, once when we designate the leftmost two students as the pair and once when we designate the rightmost two students as the pair. Since we only wish to exclude these arrangements once, we must add them back.
If all three students sit together, we have seven objects to arrange, the three empty seats, the trio of seats occupied by Alvin, John, and Albert, and the other three people. We can select three of these seven positions for the empty seats in $\binom{7}{3}$ ways. We can arrange the remaining four objects in $4!$ ways. We can arrange Alvin, John, and Albert within the designated trio of seats in $3!$ ways. Hence, the number of seating arrangements in which Alvin, John, and Albert sit together is
$$\binom{7}{3}4!3!$$
By the Inclusion-Exclusion Principle, the number of permissible seating arrangements is
$$\binom{9}{3}6! - \binom{8}{3}\binom{3}{2}5!2! + \binom{7}{3}4!3!$$
Best Answer
Since you have done the first one I will go through number $2$. For the sake of brevity I will give the highlights of the method and provide links for further explanation.
I would use rook polynomials.
In this case the chess board looks like
$$\begin{array}{cc} & \text{chairs} \\ \text{students} & \begin{array}{c|c|c|c|c|c|c|c} &2&1&3&7&13&4&\cdots \\\hline 1 &\bbox[silver,10px]{\phantom{H}} &\bbox[silver,10px]{\phantom{H}} &\bbox[silver,10px]{\phantom{H}} & & & &\cdots \\\hline 2 & &\bbox[silver,10px]{\phantom{H}} &\bbox[silver,10px]{\phantom{H}} &\bbox[silver,10px]{\phantom{H}} & & &\cdots \\\hline 7 & & & & &\bbox[silver,10px]{\phantom{H}} &\bbox[white,10px]{\phantom{H}} &\cdots \\\hline 3 &\bbox[white,10px]{\phantom{H}} & & & & & &\cdots \\\hline 4 &\bbox[white,10px]{\phantom{H}} & & & & & &\cdots \\\hline \vdots & \vdots &\vdots &\vdots &\vdots &\vdots &\vdots&\ddots \\ \end{array} \end{array}$$
We have two disjunct forbidden subboards, one consisting of a single square, this has rook polynomial
$$1+x$$
the other consists of two rows of $3$ squares with the lower $3$ offset by $1$ square. The rook polynomial for this is found by manually counting the placements of identical non-attacking rooks. There is $1$ way to place $0$ rooks, $6$ ways to place $1$ rook and $7$ ways to place $2$ rooks, hence the polynomial for this subboards is
$$1+6x+7x^2$$
since they are disjunct subboards the total rook polynomial is
$$(1+x)(1+6x+7x^2)= 1 + 7x + 13x^2 + 7x^3$$
Since the board is $N\times N$ with $N\gt 20$ (assuming the same number of students as seats) then this polynomial transforms by replacing
$$x^k\longrightarrow (-1)^k(N-k)!$$
giving our final answer:
Please see my answer here for more on rook polynomials.