How to compute that the unit digit of $\frac{(7^{6002} − 1)}{2}$? The mother problem is:

Find the unit digit in LCM of $7^{3001} − 1$ and $7^{3001} + 1$

This problem comes with four options to choose the correct answer from,my approach,as the two number are two consecutive even numbers hence the required LCM is  $$\frac{(7^{3001} − 1)(7^{3001} + 1)}{2}$$
Using algebra that expression becomes $\frac{(7^{6002} − 1)}{2}$,now it is not hard to see that unit digit of $(7^{6002} − 1)$ is $8$.
So the possible unit digit is either $4$ or $9$,As there was no $9$ as option I selected $4$ as the unit digit which is correct but as this last part is a kind of fluke I am not sure if my approach is right or not or may be I am unable to figure out the last part how to be sure that the unit digit of $\frac{(7^{6002} − 1)}{2}$ is $4$?

We look directly at the mother problem. Exactly as in your approach, we observe that we need to evaluate
$$\frac{(7^{3001}-1)(7^{3001}+1)}{2}$$
modulo $10$.
Let our expression above be $x$.  Then $2x= (7^{3001}-1)(7^{3001}+1)$.  We will evaluate $2x$ modulo $20$.
Note that $7^{3000}$ is congruent to $1$ modulo $4$ and modulo $5$. Thus  $7^{3001} \equiv 7\pmod{20}$, and therefore 
$$2x\equiv (6)(8)\equiv 8 \pmod{20}.$$
It follows that $x\equiv 4\pmod{10}$.
