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
Method 1: We arrange the five letters B, K, K, P, R in a row, then insert the two O's and three E's in that order.
The number of ways we can arrange the five letters B, K, K, P, R in a row is $$\frac{5!}{2!}$$ where we divide by $2!$ since we can permute the two K's within a given arrangement without producing an arrangement distinguishable from that arrangement.
We now have six spaces to fill, four between successive letters and two at the ends of the row. We can either place both O's in one of these six spaces, which can be done in $\binom{6}{1}$ ways, or place them in two different spaces, which can be done in $\binom{6}{2}$ ways.
Case 1: We place both O's in one of the six spaces.
This creates an arrangement with seven letters in which the two O's are adjacent. Since we are not permitted to have two adjacent O's, we must place an E in between the two O's. We now have an arrangement of eight letters, including the sequence OEO. This creates nine spaces, seven between successive letters and two at the ends of the row. Since the E's cannot be adjacent, we must separate them by placing an E in two of the seven spaces that are not adjacent to the E we have placed between the two O's. We can do this in $\binom{7}{2}$ ways. The number of such arrangements is $$\binom{6}{1}\binom{7}{2}$$
Case 2: The two O's are placed in two different spaces.
This creates an arrangement of seven letters in which the O's are not adjacent. We have eight spaces to fill, six between successive letters and two at the ends of the row. To separate the E's, we choose three of these eight spaces in which to insert an E, which can be done in $\binom{8}{3}$ ways. The number of such arrangements is $$\binom{6}{2}\binom{8}{3}$$
Total: Since the two cases are disjoint, the number of arrangements of BOOKKEEPER in which the E's are not adjacent and the O's are not adjacent is $$\frac{5!}{2!}\left[\binom{6}{1}\binom{7}{2} + \binom{6}{2}\binom{8}{3}\right] = 57960$$
Method 2: We use the Inclusion-Exclusion Principle.
There are $$\frac{10!}{2!3!}$$ distinguishable arrangements of the word BOOKKEEPER. From these we must exclude those arrangements in which two O's or two E's are adjacent.
One pair of adjacent identical letters:
Two O's are adjacent: We have nine objects to arrange: B, OO, K, K, E, E, E, P, R. Since the two K's are indistinguishable and the three E's are indistinguishable, they can be arranged in $$\frac{9!}{2!3!}$$ distinguishable ways.
Two E's are adjacent: We have nine objects to arrange: B, O, O, K, K, EE, E, P, R. Since the two O's are indistinguishable and the two K's are indistinguishable, they can be arranged in $$\frac{9!}{2!2!}$$ distinguishable ways.
Two pairs of adjacent identical letters:
Two O's are adjacent and two E's are adjacent: We have eight objects to arrange: B, OO, K, K, EE, E, P, R. Since the two K's are indistinguishable, they can be arranged in $$\frac{8!}{2!}$$ distinguishable ways.
Two pairs of E's are adjacent: We have eight objects to arrange: B, O, O, K, K, EEE, P, R. Since the two K's are indistinguishable and the two O's are indistinguishable, they can be arranged in $$\frac{8!}{2!2!}$$ distinguishable ways.
Three pairs of adjacent identical letters:
Two O's are adjacent and two pairs of E's are adjacent: We have seven objects to arrange: B, OO, K, K, EEE, P, R. Since the two K's are indistinguishable, they can be arranged in $$\frac{7!}{2!}$$ distinguishable ways.
By the Inclusion-Exclusion Principle, the number of permissible arrangements is $$\frac{10!}{2!2!3!} - \frac{9!}{2!3!} - \frac{9!}{2!2!} + \frac{8!}{2!} + \frac{8!}{2!2!} - \frac{7!}{2!} = 57960$$