$m + n = 12k_1 + 8 ------------------(1)$
$m - n = 12k_2 + 6 --------------- (2)$
Observe here $k_1$ will be greater than or equal to $k_2$
Adding $(1)\ and\ (2)$
$2m = 12(k_1+k_2+1) + 2$
let $k_1+k_2+1 = k_3$
$m = 6(k_3) + 1$
Similarly by subtracting $(2)$ from $(1),$
$n = 6(k_4) + 1$
Now $mn = 36(k_3)(k_4) + 6(k_3) + 6(k_4) + 1$
$= 6(k_5) + 1$
Clearly now $mn$ when divided by $6$, will leave remainder $\textbf{1}$