With the inclusion-exclusion principle, I would like to compute the following:
(Total number of permutations) - (Number of permutations with $2$ $C$'s together) - (Number of permutations with $2$ $S$'s together) + (Number of permutations with $2$ $C$'s together and $2$ $S$'s together) + (Number of permutations with $3$ $S$'s together) - (Number of permutations with $2$ $C$'s together and $3$ $S$'s together).
Total number of permutations: $\frac{7!}{3!2!}$. There are $7!$ orderings, but we divide out by the repeated letters.
Number of permutations with $2$ $C$'s: $\frac{6!}{3!}$. There are $6!$ orderings when we treat $CC$ as a single character, but we divide out by the repeated $S$'s.
Number of permutations with $S$ $S$'s: $\frac{6!}{2!}$. Put $2$ of the $S$'s together and leave the third one out of the pair. Divide out by the repeated $C$'s.
Number of permutations with $2$ $C$'s and $2$ $S$'s: $5!$. Put the $2$ $C$'s and $2$ of the $S$'s as a single letter.
Number of permutations with $3$ $S$'s: $\frac{5!}{2!}$. Put all $3$ of the $S$'s together and divide out by the $C$'s.
Number of permutations with $2$ $C$'s and $3$ $S$'s: $4!$. Put the $2$ $C$'s and $3$ $S$'s as single letters.
The phrasing of the question is atrocious. Let's assume the intended question was
How many words can be formed from the letters of the word SUCCESS if
- the two C's are together but no two of the three S's are together?
- no two adjacent letters are identical?
Your first solution to the first question is correct. The book's answer is incorrect.
As for your second solution, we need to exclude those arrangements in which consecutive S's appear. Notice that when you subtract the $5!$ arrangements in which two S's are together, you subtract arrangements in which all three S's are together twice, once when you count arrangements in which the first two of the three consecutive S's are counted as your double S and once when you count arrangements in which the last two of the three consecutive S's are counted as your double S. Consequently, when you apply the Inclusion-Exclusion Principle, you must add the $4!$ arrangements in which three consecutive S's appear so that arrangements in which three consecutive S's appear are only excluded once. That yields
$$\frac{6!}{3!} - 5! + 4!$$
which equals the answer
$$\frac{6!}{3!} - (5! - 4!)$$
you obtained when you assumed that three consecutive S's could not appear.
As for the second question, which you also solved correctly, here are two approaches.
First approach: Consider cases.
There are $3!$ arrangements of U, CC, E, where the double C is treated as a single letter. To ensure that no two identical letters are consecutive, we must insert an S between the two C's. Suppose we have an arrangement of the form
$$UCSCE$$
To ensure that the S's are separated, we must place the two remaining S's in one of the four spaces indicated by a wedge
$$\wedge U \wedge C S C \wedge E \wedge$$
There are
$$3!\binom{4}{2} = 36$$
distinguishable arrangements of the letters of SUCCESS in which no two consecutive letters are identical and the two C's are separated by a single S.
In the remaining arrangements, the C's must be separated by a letter other than a single S. There are $2!$ arrangements of the letters U and E, which creates three gaps in which we can place the two C's.
$$\wedge U \wedge E \wedge$$
Since we can fill two of these three gaps in $\binom{3}{2}$ ways, there are
$$2! \cdot \binom{3}{2} = 3!$$
arrangements of U, C, C, E in which the two C's are separated. Suppose we have an arrangement such as
$$C U C E$$
To ensure the three S's are also separated, we must place an S in three of the five locations indicated by a wedge
$$\wedge C \wedge U \wedge C \wedge E \wedge$$
Hence, there are
$$2!\binom{3}{2}\binom{5}{3} = 60$$
arrangements of the letters of SUCCESS in which no two letters are identical and the C's are separated by a letter other than a single S.
Since the cases are disjoint, there are
$$3!\binom{4}{2} + 2!\binom{3}{2}\binom{5}{3} = 36 + 60 = 96$$
arrangements of the letters of the word SUCCESS in which no two consecutive letters are identical.
Second approach: We can apply the Inclusion-Exclusion Principle by considering pairs of identical letters. The total number of distinguishable arrangements of the seven letters of the word SUCCESS is
$$\frac{7!}{3!2!}$$
where the $3!$ in the denominator represents the number of ways we could permute the three S's within a given arrangement without producing an arrangement that is distinguishable from the given arrangement and the $2!$ in the denominator represents the number of ways we could permute the two C's within a given arrangement without producing an arrangement that is distinguishable from the given arrangement.
We now exclude arrangements in which pairs of identical letters are adjacent.
One pair of adjacent identical letters
A pair of consecutive C's: Treat the double C as a single letter, so we are arranging the six objects S, S, S, U, E, CC, which can be done in
$$\frac{6!}{3!}$$
distinguishable ways.
A pair of consecutive S's: Treat a double S as a single letter, so we are arranging the six objects SS, S, U, E, C, C, which can be done in $$\frac{6!}{2!}$$
distinguishable ways.
Two pairs of identical letters
A pair of consecutive C's and a pair of consecutive S's: We must arrange CC, SS, S, U, E, which can be done in $$5!$$ ways.
Two pair of consecutive S's: Since there are only three S's in SUCCESS, the two pairs of consecutive S's must be a block of three consecutive S's. Thus, we must arrange the five objects SSS, C, C, U, E, which can be done in
$$\frac{5!}{2!}$$
distinguishable ways.
Three pairs of consecutive identical letters
A pair of consecutive C's and two pairs of consecutive S's: This can only occur if there is a double C and a triple S, so we must arrange the four objects CC, SSS, U, E, which can be done in
$$4!$$
distinguishable ways.
By the Inclusion-Exclusion Principle, the number of distinguishable arrangements of the letters of the word SUCCESS in which no two consecutive letters are identical is
$$\frac{7!}{2!3!} - \frac{6!}{3!} - \frac{6!}{2!} + 5! + \frac{5!}{2!} - 4! = 96$$
The answer to which you linked requires a much deeper knowledge of mathematics than that described in your post.
Best Answer
Ignore the C's for the moment and arrange the remaining letters. There are $\frac{12!}{5!3!2!}=332640$ ways to do this.
Now consider the thirteen spaces between and beyond the letters – at most one C may be inserted into each space, so given an arrangement of the other letters there are $\binom{13}3=286$ ways to add C's.
The total number of admissible combinations is thus $332640×286=95135040$.