Your answer to the first question is correct.
How many permutations of $a, a, b, b, b, c, c, c, c$ are there such that no three consecutive elements are the same?
The total number of arrangements is
$$\binom{9}{2}\binom{7}{3}\binom{4}{4} = \frac{9!}{2!7!} \cdot \frac{7!}{3!4!} \cdot \frac{4!}{4!0!} = \frac{9!}{2!3!4!0!} = \frac{9!}{2!3!4!}$$
since we must choose two of the nine positions for the $a$s, three of the remaining seven positions for the $b$s, and fill all four of the remaining four positions with the $c$s. We can express the number of elements in the sample space using a multinomial coefficient
$$\binom{9}{2, 3, 4} = \frac{9!}{2!3!4!}$$
From these, we must subtract those arrangements in which there is at least one block of three consecutive identical letters.
A block of three consecutive identical letters: Since there are only two $a$s, this can only happen in two ways, either there is a block of three consecutive $b$s or a block of three consecutive $c$s.
There is a block of three consecutive $b$s: Then we have seven objects to arrange: $a, a, bbb, c, c, c, c$. Choose two of the seven positions for the $a$s, one of the remaining five positions for the block, and fill all four of the remaining four positions with the $c$s, which can be done in
$$\binom{7}{2, 1, 4}$$
ways.
There is a block of three consecutive $c$s: Then we have seven objects to arrange: $a, a, b, b, b, ccc, c$. Choose two of the seven positions for the $a$s, three of the remaining five positions for the $b$s, one of the remaining two positions for the block of three $c$s, and fill the remaining position with the remaining $c$. There are
$$\binom{7}{2, 3, 1, 1}$$
such arrangements.
Two blocks, each containing three consecutive identical letters: There are two ways this can occur, either there are three consecutive $b$s and three consecutive $c$s or there are four consecutive $c$s (two overlapping blocks of three consecutive $c$s, the first three and the last three).
A block of three consecutive $b$s and a block of three consecutive $c$s: We have five objects to arrange: $a, a, bbb, ccc, c$. Choose two of the five positions for the $a$s, one of the remaining three positions for the block of $b$s, one of the remaining two positions for the block of $c$s, and fill the remaining position with the remaining $c$. There are
$$\binom{5}{2, 1, 1, 1}$$
such arrangements. It looks like you counted this case twice in $S_2$, so you added too much to your answer.
A block of four consecutive $c$s: We have six objects to arrange: $a, a, b, b, b, cccc$. Choose two of the six positions for the $a$s, three of the remaining five positions for the $b$s, and fill the remaining position with the block of $c$s, which can be done in
$$\binom{6}{2, 3, 1}$$
ways.
Three blocks, each containing three consecutive identical letters: For this to occur, we must have three consecutive $b$s and four consecutive $c$s. Hence, we have four objects to arrange: $a, a, bbb, cccc$. Choose two of the four positions for the $b$s, one of the remaining two positions for the block of $b$s, and then fill the remaining position with the block of $c$s. There are
$$\binom{4}{2, 1, 1}$$
such arrangements.
By the Inclusion-Exclusion Principle, the number of arrangements of $a, a, b, b, b, c, c, c, c$ in which no block of three consecutive identical letters appears is
$$\binom{9}{2, 3, 4} - \binom{7}{2, 1, 4} - \binom{7}{2, 3, 1, 1} + \binom{5}{2, 1, 1, 1} - \binom{6}{2, 3, 1} + \binom{4}{2, 1, 1}$$
What mistakes did you make?
All of your counts match mine except a block of three consecutive $c$s, which you subtracted twice, and a block of three consecutive $b$s and three consecutive $c$s, which you added twice. When you have four $c$s and three of them are consecutive, you must treat the four $c$s as two objects: $ccc$ and $c$. The two objects may or may not be adjacent. If they are adjacent, they form two blocks of three consecutive $c$s: $\color{red}{ccc}c$ and $c\color{red}{ccc}$.
How many permutations of $a, a, b, b, b, c, c, c, c$ are there such no that two consecutive letters are the same.
When you attempt this question, keep in mind that three consecutive $b$s count as two pairs of adjacent $b$s, three consecutive $c$s count as two pairs of consecutive $c$s, four consecutive $c$s count as three pairs of consecutive $c$s, and that you can also have two disjoint pairs of adjacent $c$s, which may or may not be adjacent.
You may want to reread my answer to the linked question before you attempt this question.
