How to solve these types of permutation problems? One thing I observe that, the way we calculate $nPn$, doesn't actually work for the cases of $nPr$.
(1) How many permutations are possible if we take 3 letters at a time from the list {A,A,B,C,E} where the words always start with an A and ends with an A?
Ans:  $\frac{^{5-2}P_{3-2}}{2!} = 3$ .
(2) How many permutations are possible if we take 3 letters at a time from the list {A,A,B,C,E} where the words always start with an AA?
Ans:  $\frac{^{5-2}P_{3-2}}{2!} = 3$ .
(3) How many permutations are possible if we take 3 letters at a time from the list {A,A,B,C,E} where the words always start with an A?
Ans:  $^{5-1}P_{3-1} = 12$ .
(4) How many of them contain AA?
Answer: 3P2 = 6.
(5) How many permutations are possible if we take 3 letters at a time from the list {A,A,B,C,E} where the words always start with an E?
Ans: 
The answer should be 7. 
But, $^{5-1}P_{3-1} = 12$ is wrong.
(6) How many of them start with a vowel?
Ans: 
The answer should be 19. 
But, $^{5-3}P_{3-3} = 2$ is wrong.
My question is, which problems should I solve by using multiplications between different permuted values, which questions should I solve by using additions between different permuted values, and, which problems should I solve by hand-counting (ie. Permutation formula doesn't work)?
Another question is, how should I deal with overcounting?
 A: The answer to 5 th is 7 . See  we have to select three lettets where first letter E is fixed. Now  from  4 letters we have to select two so we have  4C1 but A is common so divided by 2 and next letter can be selected in 3C1 ways . So answer is 1+(4C1/2.3C1)=7. Note  the difference between permutation and combination. Combination is selection of n things taken r at a time while permutation is arrangement of distinct things taken r at a time.
A: Try choosing the most restrictive elements you can, then identify a subproblem to solve.
E.g. to count the permutations of the alphabet that starts with A and ends with Z, I know I must place A and Z at either end, so the answer will be the ways to permute the 24 remaining letters in the middle, $24!$.
One way strategy might be to count the ways you can construct a new set for which the direct permutation calculation can be applied.
E.g. to count the 4-permutations of $\left\{1,2,3,4,5,6\right\}$ that contain a $1$ and a $4$, form a new set by picking $1$ and $4$ and 2 other elements. For this new set there are $4!$ permutations. But there were $^4C_2$ ways to choose those extra elements, so the answer is $4! \times ^4C_2$.
Edit:
(when I say n-permutation I mean a permutation of the set containing n items. I think this is standard terminology)
E.g. find 5-permutations of $\left\{1,1,1,2,3,4\right\}$ that have 3 $1$s next to each other. As before: deal with the difficult stuff first. As we have to have the $1$s next to each other, think of them as a single element. So now the problem becomes "find the 3-permutations of $\left\{\left\{1,1,1\right\},2,3,4\right\}$ that contains the $\left\{1,1,1\right\}$". Dealing with the difficult stuff first, we know we must include the $1$s. Form a new set by choosing the $1$s and 2 other elements. There are $3!$ ways to permute this new set. But there were $^3C_2$ ways to choose those extra elements, so the answer is $3! \times ^3C_2$.
When the set contains multiple copies of an element (a multiset) the reasoning that gives the $^nP_r$ formula no longer holds. Some permutations may contain several copies of an element and so are overcounted, some aren't. You can transform the problem into something more familiar as I have above, or perhaps count the cases of overcounting separately, adjust for it, then add up your separate cases.
E.g. find the 3-permutations of $\left\{A,A,B,C,D\right\}$. Counting the permutations with 2 $A$s, there are $^3C_1 = 3$ ways to choose the 3rd element after being forced to choose the 2 $A$s. There are $3!$ ways to permute this new set. But we have counted each permutation twice, so divide by $2$, = $\frac{3 \times 3!}{2} = 9$. With just 1 $A$, there are $^3C_2 = 3$ ways to choose the remaining 2 elements, and $3!$ ways to permute, with no overcounting this comes to $3 \times 3! = 18$. With no $A$s there are $^3C_3 = 1$ way to choose the 3 elements, with $3!$ permutations, with no overcounting this is $6$ permutations. So the answer is $9 + 18 + 6 = 33$.
A: Using intuition and not formulae
I would try to use common sense first in solving counting problems. Whether it is a permutation or a combination is just incidental. 
To take your example, Question (5). The 3-letter word starts with $E$. Now there are $3$ cases on the other 2 letters (a) No $A$ is used ... $\color{blue}{2}$ arrangements (b) Exactly one $A$ is used ... $\color{blue}{4}$ arrangements and (c) Both $A$'s are used ... $\color{blue}{1}$ arrangement. So the total is $2 + 4 + 1 = \color{blue}{7}$ arrangements. Here the question of using a formula did not even arise.
Question (6).  We want vowel first, so there are $2$ cases (a) The word starts with $A$. which leaves us $\{A, B, C, E \}$ to choose 2 letters, that is $4P2$ which is $\color{blue}{12}$ ways and (b) The word starts with $E$ which leaves us $\{A, A, B, C \}$ to choose 2 letters. The repetition of the A's is dealt similarly as in Question (5) and we will get $1 + 6 = \color{blue}{7}$ ways. Hence you get $12 + 7 = \color{blue}{19}$ as your answer.
On your question of not double counting, once you start using common sense reasoning and not just formulae, you will probably be pretty safe. 
