I think you're on the right track, but I wouldn't recommend fixing the first 3 spots to be distinct numbers as you do. For example, the number $11231$ satisfies the condition of being a 5-digit number with exactly $3$ different digits, but doesn't have the first $3$ digits distinct.
I noticed that you have $10 * 9 * 8 = 720$ ways to pick your first $3$ digits to use in order. Instead, I would suggest that you just find the total number of combinations of $3$ digits you could use in making your string, which is $10$ choose $3$, or $120$. Given this, the next step would be to split the problem into cases. As you seem to have noticed, you can either have $2$ of the digits be used twice, or $1$ of the digits be used thrice.
Let's consider case $1$, which is that you have $2$ digits used twice. There's $3$ ways to choose which of the $3$ digits to use twice, and then a total of $\frac{5!}{2! * 2!} = 30$ ways to arrange the digits afterwards, giving us a total of $120 * 3 * 30 = 10800$ possible arrangements.
Now, try to calculate the other case by yourself and see if you're able to arrive at the answer.
This question is kinda weird since you didn't say anything about repetition. Supposing our sequence of length n allows repetition, here is how I would approach it:
We want to create sequences of length n from a set of $k$ characters allowing repetition with the constraint that at most two adjacent characters can be the same.
Let's use the inclusion-exclusion principle.
First, the number of sequences of length $n$ from $k$ elements allowing repetition is: $k^n$.
Now let's define the following sets for $ 3 \leq p \leq n$:
$$
S_p=\{\text{Sequences of length n with p adjacent characters the same}\}
$$
Therefore our final answer is: $k^n- |\bigcup_{p=3}^n S_p|$. Let's then calculate the cardinality of that union.
Clearly those sets are all disjoint, therefore:
$$
|\bigcup_{p=3}^n S_p| = |S_3| + |S_4| + \cdots + |S_n|
$$
And we can use the following strategy to calculate the cardinality of $|S_p|$: From our $k$ characters, select $1$ and place it in $p$ adjacent positions. There are $n-p+1$ possibilities to choose adjacent positions. Now we need to let the remaining $n-p$ positions to select, without repetition, characters from the $k-1$ remaining, and that can be done in $(k-1)(k-2)\cdots (k-n+p)=\frac{(k-1)!}{(k-n+p-1)!}$ ways. Therefore, by multiplication principle we get:
$$
|S_p|=k\cdot (n-p+1) \cdot \frac{(k-1)!}{(k-n+p-1)!} = \frac{k!(n-p+1)}{(k-n+p-1)!}
$$
which follows that:
$$
|\bigcup_{p=3}^n S_p| = \sum_{p=3}^{n} \frac{k!(n-p+1)}{(k-n+p-1)!}
$$
Finally, our answer is:
$$
k^n - \sum_{p=3}^{n} \frac{k!(n-p+1)}{(k-n+p-1)!}
$$
Sanity Check: Suppose $n=4$ and $K=\{a,b\}$. Therefore all sequences are:
\begin{align*}
(a,a,a,a)&\\
(a,a,b,b)&,(a,b,a,b)\\
(a,a,a,b)&,(a,a,b,a),(a,b,a,a),(b,a,a,a)\\
(b,b,b,b)&\\
(b,b,a,a)&,(b,a,b,a)\\
(b,b,b,a)&,(b,b,a,b),(b,a,b,b),(a,b,b,b)\\
(b,a,a,b)&,(a,b,b,a)
\end{align*}
There are $2^4=16$ sequences. Removing the ones that we don't want to count, that is, sequences with three or more adjacent repetitions, we have the remaining sequences:
$$
(a,a,b,b),(a,b,a,b)\\
(a,a,b,a),(a,b,a,a)\\
(b,b,a,a),(b,a,b,a)\\
(b,b,a,b),(b,a,b,b)\\
(b,a,a,b),(a,b,b,a)
$$
There are $10$ sequences. Now let's check if our answer is going to validate to the same number:
\begin{align*}
k^n - \sum_{p=3}^{n} \frac{k!(n-p+1)}{(k-n+p-1)!} &= 2^4 - \frac{2!(4-3+1)}{(2-4+3-1)!} - \frac{2!(4-4+1)}{(2-4+4-1)!}\\
&= 16 - \frac{2!(2)}{(0)!} - \frac{2!(1)}{(1)!}\\
&= 16 - 4 - 2 = 10
\end{align*}
So it seems to make sense. I'm not sure if it's 100% correct, but that's what I've done to try to solve it. Hope it helps!
Best Answer
In your cases $II, III, IV$ you have overcounted because swapping the $@$ signs results in the same string. So in case $II$, instead of $5 \cdot 4$ ways to choose where the $@$ signs are, there are $5 \choose 2$ The fact that you are close supports that there are $128$ total ASCII characters
Less work is to compute how many total strings there are and subtract the ones with no $@$. $128^5-127^5=1,321,368,961$