Multiplied by 6 digits, the mantissa is 0, so the digits can only be 0 or 5; But after the multiplicand digit is multiplied by the multiplier digit, the mantissa is 5, so the multiplicand digit cannot be 0, only 5.

Multiplied by 6, the product is greater than 1000 and less than 2000, so the hundredth of the multiplicand is 2, that is, the multiplicand is 245.

After multiplying the multiplier's ten digits by the multiplicand's single digit 5, the mantissa is 5, so the multiplier's ten digits can only be odd, and the highest digit after multiplying by the multiplicand has no carry, so it can only be 3 (if it is 1, the product is much less than 8000), so the first question should be 245x36=8820.

The 10 digit of the multiplicand is 7. After multiplying by the multiplier 4, its 100 bit has a carry 2 or 3. The 100 digit of the multiplicand multiplied by the 4 digits of the multiplier must be an even number, and after adding the carry, it is equal to 4, so the carry must be an even number 2, that is, the result of multiplying the 4 digits of the multiplicand cannot have a carry greater than 1. So the single digit of the multiplicand must be 2, not 6.

The decimal places of the multiplier are multiplied by the multiplicand digits 2, and the mantissa is 2, so the decimal places of the multiplier are 1 or 6. However, after the decimal places of the multiplier are multiplied by the multiplicand, the decimal places of the multiplier have a carry from 2 to thousands, so the decimal places of the multiplier cannot be 1, but only 6.

After the multiplicand is multiplied by the ten digits of the multiplier, the result is 22x2, so the hundred digits of the multiplicand should be 3.

The answer to the second question should be 372x64= 1488+22320=23808.

It takes too much time to write instructions. I hope you can see clearly. I feel that this problem is a bit troublesome for the fourth grade.