Solution:

Let $T_{n}=a^{n}+b^{n}+c^{n}$ and consider the polynomial
$$P(x)=(x-a)(x-b)(x-c)=x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c .$$
Since $P(a)=0$, we get $a^{3}=(a+b+c) a^{2}-(a b+a c+b c) a+a b c$ and multiplying both sides by $a^{n-3}$ we obtain $a^{n}=(a+b+c) a^{n-1}-(a b+a c+b c) a^{n-2}+(a b c) a^{n-3}$. Applying the same reasoning, we can obtain similar expressions for $b^{n}$ and $c^{n}$ and adding the three identities we get that $T_{n}$ satisfies the following 3-term recurrence:
$$T_{n}=(a+b+c) T_{n-1}-(a b+a c+b c) T_{n-2}+(a b c) T_{n-3} \text{, for all } n \geq 3$$
From this we see that if $T_{n-2}$ and $T_{n-3}$ are divisible by $a+b+c$, then so is $T_{n}$. This immediately resolves part (b) - there are no ordered triples which are 2004-powerful and 2005-powerful, but not 2007-powerful - and reduces the number of cases to be considered in part (a): since all triples are 1-powerful, the recurrence implies that any ordered triple which is both 2-powerful and 3-powerful is $n$-powerful for all $n \geq 1$.

Putting $n=3$ in the recurrence, we have
$$a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)-(a b+a c+b c)(a+b+c)+3 a b c$$
which implies that $(a, b, c)$ is 3-powerful if and only if $3 a b c$ is divisible by $a+b+c$. Since
$$a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+a c+b c),$$
$(a, b, c)$ is 2-powerful if and only if $2(a b+a c+b c)$ is divisible by $a+b+c$.

Suppose a prime $p \geq 5$ divides $a+b+c$. Then $p$ divides $a b c$. Since $\operatorname{gcd}(a, b, c)=1$, $p$ divides exactly one of $a, b$ or $c$; but then $p$ doesn't divide $2(a b+a c+b c)$.

Suppose $3^{2}$ divides $a+b+c$. Then 3 divides $a b c$, implying 3 divides exactly one of $a$, $b$ or $c$. But then 3 doesn't divide $2(a b+a c+b c)$.

Suppose $2^{2}$ divides $a+b+c$. Then 4 divides $a b c$. Since $\operatorname{gcd}(a, b, c)=1$, at most one of $a, b$ or $c$ is even, implying one of $a, b, c$ is divisible by 4 and the others are odd. But then $a b+a c+b c$ is odd and 4 doesn't divide $2(a b+a c+b c)$.

So if $(a, b, c)$ is 2- and 3-powerful, then $a+b+c$ is not divisible by 4 or 9 or any prime greater than 3. Since $a+b+c$ is at least $3$, $a+b+c$ is either 3 or 6. It is now a simple matter to check the possibilities and conclude that the only triples which are $n$-powerful for all $n \geq 1$ are $(1,1,1)$ and $(1,1,4)$.

Let $p$ be a prime. By Fermat's Little Theorem,
$$a^{p-1} \equiv \begin{cases}1(\bmod p), & \text{ if } p \text{ doesn't divide } a \\ 0(\bmod p), & \text{ if } p \text{ divides } a\end{cases}$$
Since $\operatorname{gcd}(a, b, c)=1$, we have that $a^{p-1}+b^{p-1}+c^{p-1} \equiv 1,2$ or $3(\bmod p)$. Therefore if $p$ is a prime divisor of $a^{p-1}+b^{p-1}+c^{p-1}$, then $p$ equals 2 or 3. So if $(a, b, c)$ is $n$-powerful for all $n \geq 1$, then the only primes which can divide $a+b+c$ are 2 or 3.

We can proceed in a similar fashion to show that $a+b+c$ is not divisible by 4 or 9.

Since
$$a^{2} \equiv \begin{cases}0(\bmod 4), & \text{ if } a \text{ is even;} \\ 1(\bmod 4), & \text{ if } a \text{ is odd }\end{cases}$$
and $a, b, c$ aren't all even, we have that $a^{2}+b^{2}+c^{2} \equiv 1,2$ or $3(\bmod 4)$.

By expanding $(3 k)^{3},(3 k+1)^{3}$ and $(3 k+2)^{3}$, we find that $a^{3}$ is congruent to 0, 1 or $-1$ modulo 9. Hence
$$a^{6} \equiv \begin{cases}0(\bmod 9), & \text{ if } 3 \text{ divides } a ; \\ 1(\bmod 9), & \text{ if } 3 \text{ doesn't divide } a .\end{cases}$$
Since $a, b, c$ aren't all divisible by 3, we have that $a^{6}+b^{6}+c^{6} \equiv 1,2$ or $3(\bmod 9)$.

So $a^{2}+b^{2}+c^{2}$ is not divisible by 4 and $a^{6}+b^{6}+c^{6}$ is not divisible by 9. Thus if $(a, b, c)$ is $n$-powerful for all $n \geq 1$, then $a+b+c$ is not divisible by 4 or 9. Therefore $a+b+c$ is either 3 or 6 and checking all possibilities, we conclude that the only triples which are $n$-powerful for all $n \geq 1$ are $(1,1,1)$ and $(1,1,4)$.

See Solution for the (b) part.