$3^4=81 \equiv 1 \bmod 10\Rightarrow (3^4)^{23} \equiv 1 \bmod 10\Rightarrow 3^{92}\equiv 1 \bmod 10\Rightarrow(3^{92})^{94} \equiv 1 \bmod 10.$
$7^2=49 \equiv -1 \bmod 10\Rightarrow (7^2)^{1005} \equiv -1 \bmod 10\Rightarrow 7^{2010}\equiv -1 \bmod 10\Rightarrow(7^{2010})^{2012} \equiv 1 \bmod 10.$
Vậy
$(7^{2010})^{2012} - (3^{92})^{94} \equiv 0 \bmod 10\Rightarrow $ đpcm.