2. 1.) let a=0, then 0.b=0.c, so b=0, c=1 and b=1, c=0 are the two solutions such that $b\ne c$. (if a=1, then 1.b=1.c, so b=c)
2.) using a=0 from (1), a+b=a+c $\implies 0+b=0+c \therefore b=c$