Note first that it may help to factorise your large number if it is not prime. Here there is a fairly obvious factor $7$ ($203=210-7$) and you get $223321= 7 \times 31903$.
Now look at the Fibonacci Sequence modulo $7$, which goes $1,1,2,3,5,1,6,0,6,6,5,4,2,6,1,0,1,1 \dots$.
It is then easy to show that every eighth term is divisible by $7$. You will have to do less work if you know the theorem that if $F_1=F_2=1; F_{n+1}=F_n+F_{n-1}$ then $F_n\mid F_{kn}$.
Then you can use the same method with the second factor, which turns out to be $61 \times 523$. Then you know that the position in the sequence is divisible by $8$ and two other numbers - take the least common multiple to find the index.
So this gives lcm of $8, 15, 524$ which is $15720$
$1024240=2^4\times 5 \times 7 \times 31 \times 59$ so the index can be found a great deal more quickly using the factorisation.
So $16$ is a factor every $12^{th}$ number, $5$ gives $5$, $7$ gives $8$, $31$ gives $30$ and $59$ gives $58$.
So we need the least common multiple of $12,5,8,30,58$ - this makes it the $3480^{th}$ element of the sequence
Lemma $11$ of this paper shows that an odd prime divides one of $F_{p-1}, F_p, F_{p+1}$ and tells you which, and the paper gives information on how to compute the index when the prime factorisation is known.
To summarise $5\mid F_5$, and the option is $p-1$ where $p$ is congruent to $\pm 1 \bmod 5$, $p+1$ otherwise. But this need not be the least index for a prime, and doesn't deal with prime powers.
You are wrong in assuming that even numbers can't be divisible by 3. As a counter example, 6 is divisible by 3.
The divisibility rule for 3 states that the if the sum of the digits of the number is divisible by 3, the number itself is divisible 3.
Out of the seven digits given, you need to find groups of 5 for which the sum is divisible by 3. For example, $(1, 2, 3, 4, 8)$ is one such group. For every such group you need to find the number of 5 digit numbers that can be formed and get the total.
Best Answer
For a number to be divisible by $4$, the last 2 digits have to form a 2-digit number that is divisible by $4$. This should simplify things a lot.
The trick for $11$: you already know.
And if $ABCD$ is divisible by both $4$ and $11$, it is divisible by $44$.