Solution

1. "For every a in X, a*a = a" - TRUE
   Proof - Let a*a = b.
   => (a * a) * a = b * a
   => Since (A, *) is a semigroup, * is closed and associative.
   => So, (a * a) * a = a * (a * a)
   => a * b = b * a, which is possible only if a = b.
   => Hence, a * a = a.
    
2. For every a, b in X, a*b*a= a - TRUE
   Proof - Let (a * b) * a = c
   => ((a * b) * a) * a = c * a
   => (a * b) * (a * a) = c * a
   => (a * b) * a = c * a
   => c * a = a ----- (1)
   Similarly, a * (a + b + a) = a * c
   => a * (a * (b * a)) = a * c
   => (a * a) * (b * a) = a * c
   => a * (b * a) = a * c
   => (a * b) * a = a * c
   => c = a * c ----- (2)
   From (1) and (2), we can get c * a = a * c => c = a.
    
3. For every a, b, c in X, a*b*c= a*c - TRUE
   Proof - Let (a * b) * c = d
   => (a * b) * c * c = d * c
   => a * b * c = d * c
   => d * c = d.
   Similarly, a * (a * b * c) = a * d
   => a * a * (b * c) = a * d
   => a * (b * c) = a * d
   => a * d = d.
   Thus, d * c = a * d = d.
   => Now, c * d * c = c * a * d = c * d
   => c = c * a * d = c * d
   and d * c * a = a * d * a = d * a
   => d * c * a = a = d * a
   Hence, a * c = (d * a) * (c * d)
   => d * (a * c) * d = d.
   Therefore, a * b * c = a * c.