How many arrangements are there of the word POISONS so that no two vowels are together? I actually have 3 questions to ask. You just have to say if my solution is right or not.
First question:
How many arrangements are there of the word POISONS so that no two vowels are together?
I tried:
$\frac {5!}{2!} * C(6, 3) * \frac {3!}{2!}$  
Second question:
How many ways are there to place 65 different fish into 5 bowls so that each bowl has exactly 13 fish?
I tried:
P(65, 13) * P(52, 13) * P(39, 13) * P(26, 13) * P(13, 13)
Third question:
How many ways are there to distribute 13 identical blue balls and 2 identical red balls into 5 boxes with no box containing more than 1 red ball?
I tried:
C(5, 2) * C(17, 4)
 A: For the first question:
First place the consonants in the following way $$-P-S-N-S-,$$ so, this can be done in  $$4!/2!=12$$ ways. Now place the vowels in the gaps, so this can be done in $$\frac{\binom 53 \times 3!}2=30$$ ways. So, the total number of way is $$\bbox[border:2px solid red]
{12\times 30=360}$$ ways.
Note: here is similar answer.
For the second one:
The order of the fishes is irrelevant, so, answer is $$\bbox[border:2px solid red]
{\binom {65}{13}\times\binom {52}{13}\times\binom {39}{13}\times\binom {23}{13}\times\binom {13}{13}\\=\frac{65!}{(13!)^5}}$$
For the third one:
Here I assume that the boxes are distinct. So, you have only to count the number of blue balls in one box. 
Let, $b_1,b_2,b_3,b_4,b_5$ be the three box, and define $x_i$ as the number of balls in the $i^{th}$ box. So, you have to count the number of solutions to the equation $$\sum_{i=1}^5 x_i=13,$$ which is $$\binom{17}{4}.$$ And now choose any $2$ box out of $5$ to keep the $2$ red balls, in $\binom 52$ ways, so, number of ways is $$\bbox[border:2px solid red]
{\binom{17}{4}\times \binom 52}$$
A: For your second question, what you missed is that one would not care about the order of the fish in a specific bowl (they will after all swim around and mix that order up anyway).  So instead of permutations you should be using combinations, and the answer is 
$$\binom{65}{13} \binom{52}{13} \binom{39}{13} \binom{26}{13} \binom{13}{13}$$
For your third question, assuming the boxes are distinguishable, you have one part of the problem right:  You have $\binom{13+4}{4}=\binom{17}{4}$ ways to distribute the 13 blue balls among the boxes.  You then have to independently distribute the 2 red balls, which you can do in $\binom{2+4}{4}=\binom{6}{2}$ ways.  So the answer is 
$$\binom{6}{2}\binom{17}{4}$$
and your answer is correct in the limit as $5\to 6$. (LOL)
Your first question is actually a lot harder.  First you have to answer the "hard part" question "In how many ways can I arrange three arbitrary vowels and four arbitrary consonants such that no two vowels are consecutive."  IN the end you will multiply that answer by $12$ ways to permute the 4 consonants with two of them identical, times $3$ ways to permute the 3 vowels with two of them identical.  
To do the hard part , break the problem up into four cases: Either (a) the first and last letters are vowels, or (b) the first and last letters are  consonants, or (c) the first is a vowel and the last is a consonant, or (d) the first is a consonant and the last is a vowel. 
In case (a), each vowel other than the first must be preceded by a consonant, so we have to place a consonant in position 6, and we can "glue" a consonant in front of the first vowel.  We then have to place two consonants and one c-v pair into three possible slots; there are 3 ways to do this.
In case (b), there is only one valid arrangement, with the consonants and vowels alternating. In case (c) each of the remaining 2 vowels must be preceded by a consonant, so we have to arrange 2 c-v pairs and one loos consonant into 3 positions, which is done in 3 ways.  And case (d) has, by symmetry, the same answer as case (c).  The total is 10, so the answer to the first question is $$12 \cdot 3 \cdot 10 = 360$$.
