From the given number 975312468, we need to form a number divisible by 33 by adding one more digit, which can be inserted between any two existing digits, before the first, or after the last. Is this possible?
A number is divisible by 33 - which is $3 \cdot 11$, and here 3 and 11 are relatively prime - if and only if it is divisible by both 3 and 11. The necessary and sufficient condition for divisibility by 3 is that the sum of the digits is a multiple of 3. The sum of the digits of the given number is divisible by 3, so the digit to be appended can only be $0, 3, 6$ or 9.

A number is divisible by 11 if and only if the difference between the sum of the digits in the odd positions (counting from the right) and the sum of the digits in the even positions is a multiple of 11. This difference can also be formed by taking the digits from right to left with alternating + and - signs and summing them; in our case:

$$
8-6+4-2+1-3+5-7+9=9
$$

Thus, our given number is not divisible by 11. (It is also true that the remainder when divided by 11 is the same as the obtained sum, which is 9.)

From this, we immediately see that adding a 9 at the beginning of the number ensures divisibility by 11, since the next sign in the sequence is minus, and $9-9=0=0 \cdot 11$. The found 9 also ensures divisibility by 3, so one valid number is:

$$
\begin{array}{llllllllll}
9 & 9 & 7 & 5 & 3 & 1 & 2 & 4 & 6 & 8 .
\end{array}
$$

This can also be viewed as inserting a 9 between the first two digits of our given number (from the left). No other digit can fit in these positions, as there are no two digits whose difference is 11.

If we insert the digit $x$ at the far right, the alternating sum becomes $x-9$, since (1) the sign of each term on the left side of the equation changes, so the same applies to the sum. Here again, only $x=9$ makes the sum divisible by 11, so another solution is:

# $\begin{array}{lllllllllll}9 & 7 & 5 & 3 & 1 & 2 & 4 & 6 & 8 & 9 .\end{array}$ 

The examination of the remaining 7 positions can be mechanized by writing down separately for each position the sum of the digits to the right and the sum of the digits to the left (the latter starting with a minus sign, because with the insertion of a digit, the position number of the leftmost digit becomes 10, which is even) - and then only the change caused by the new or omitted digit needs to be considered - and then finding the digit $x$ that complements the sum of the two sums to 0, $\pm 11$, $\pm 22$... 

Between the 7 and the 5: $(-9+7)-x+(8-6+4-2+1-3+5)$

$$
\text { in short: }(-2)-x+(+7)=-x+5, \quad x=5
$$

Between the 5 and the 3: $\quad(-7)+x+(+2)=x-5, \quad x=5$;

(the sum of the digits to the left changed by -5, and the sum of the digits to the right also changed by -5, because the +5 term at the end was omitted)

| Between the 3 and the 1: $(-4)-x+(+5)=-x+1$, | $x=1 ;$ |
| :--- | :--- |
| Between the 1 and the 2: $(-5)+x+(+4)=x-1$, | $x=1 ;$ |
| Between the 2 and the 4: $(-3)-x+(+6)=-x+3$, | $x=3 ;$ |
| Between the 4 and the 6: $(-7)+x+(+2)=x-5$, | $x=5 ;$ |
| Between the 6 and the 8: $(-1)-x+(+8)=-x+7$, | $x=7$. |

According to these, only the 3 inserted between the 2 and the 4 makes the given number divisible by 33:

$$
\begin{array}{llllllllll}
9 & 7 & 5 & 3 & 1 & 2 & 3 & 4 & 6 & 8 .
\end{array}
$$

In summary: with the specified insertions, 3 valid numbers can be formed in 4 different ways.