A={x|x^2-4x+3=0}
={x| (x-1)(x-3) =0 }
= {1,3}
A∪B=A
=> B is a subset of A
B={x|x^2-ax+a-1=0}
x=3
x^2-ax+a-1=0
9-3a+a-1=0
a=4
A∩C=C
=> C is a subset of A
C＝{x|x^2-mx+1=0}
x=1
x^2-mx+1=0
1-m+1=0
m=2
x=3
9-3m+1=0
m =10/3
ie m=2 or 10/3