Five. (Full marks 20 points) There is a 14-digit number, the digit in the units place is 3 less than the digit in the tens place, and the new four-digit number formed by reversing its digits differs from the original four-digit number by 8987. Find this four-digit number and write out the reasoning process.

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Five, let the thousands digit of this four-digit number be $a$, the hundreds digit be $b$, the tens digit be $c$, and the units digit be $c-3$.
This number is $1000a + 100b + 10c + (c-3)$.
The new four-digit number is $1000(c-3) + 100c + 10b + a$.

According to the problem, we have
$$
\begin{array}{l}
1001(a+c-3) + 110(b+c) = 8987 . \\
1001(a+c) + 110(b+c) = 11990 . \\
91(a+c) + 10(b+c) = 1090 .
\end{array}
$$

The right side of the equation has a factor of 10, the second term on the left side has a factor of 10, and the first term $91(a+c)$ on the left side should also have a factor of 10. Since $91 = 7 \times 13$, it follows that $a+c$ is a multiple of 10.
Also, $1 \leqslant a \leqslant 9, 4 \leqslant c \leqslant 9$,
so $5 \leqslant a+c \leqslant 18$, which gives $a+c=10$.
Substituting (2) into (1) yields $b+c=18$.
Solving, we get $a=1, b=c=9, c-3=6$.
Answer: The four-digit number is 1996.