a) Since $7n = 7^2 + 7^3 + 7^4 + \ldots + 7^{2017} + 7^{2018}$, we have $6n = 7n - n = 7^{2018} - 7$. It follows that $7^{2018} = 6n + 7 = 6(n + 1) + 1$, and the remainder of the division by $6$ of the number $7^{2018}$ is $1$.

Also, $n = 7 + 7^2(1 + 7) + 7^4(1 + 7) + \ldots + 7^{2016}(1 + 7) = 7 + 8 \cdot (7^2 + 7^4 + \ldots + 7^{2016}) = 7 + 8p$, where $p = 7^2 + 7^4 + \ldots + 7^{2016}$. Hence, $7^{2018} = 6(8p + 7) + 7 = 48p + 42 + 7 = 48p + 49 = 48(p + 1) + 1$, wherefrom we conclude that the remainder of the division of $7^{2018}$ by $48$ is $1$.

b) We have $6n = 7^{2018} - 7$. Denoting by $u_2(x)$ the number built with the last two digits of the number $x$, we have $u_2(7^{4k}) = 01$, $u_2(7^{4k+1}) = 07$, $u_2(7^{4k+2}) = 49$, $u_2(7^{4k+3}) = 43$.

Then $u_2(7^{2018}) = u_2(7^{4 \cdot 504 + 2}) = 49$. Hence $u_2(6n) = u_2(49 - 7) = 42$.