First treat I and E as the same letter. Then, once you have decided where in the final arrangement they go, how many ways are there to distribute Is and Es to those positions, keeping in mind that the first of them must be an I?
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 $$\bbox[border:2px solid red]
{\binom{17}{4}\times \binom 52}$$
Best Answer
By PIE (principal of inclusion and exclusion), we first count the number of ways such that $n(M) + n(E) + n(R)$. This is equivalent to $\frac{7!}{2} + 7! + \frac{7!}{2} = 2 \cdot 7! = 10080$ (since in the first and third cases, we have two $E$'s, so we divide by $2$.
Now, we subtract $n(M \cap E) + n(M \cap R) + n(E \cap R)$, which is $6! + \frac{6!}{2} + 6! = 1800$. Finally we add back $n(M \cap E \cap R) = 5! = 120$, so the answer should be $10080 - 1800 + 120 = \boxed{9120}$