1.1  At first,the proof seems perfect!Horses 1 through n-1 are the same color,and horses2 through n are the same color,
but when the n=2 ,that 1 and 1 are same color, 2 and 2 are same color,so it can't induction that all horses are same color.
The basis of this inducation is wrong!It should from n=2 no n=1!
    1.2 This exercise is variant of Hanoi Tower.
        T(0)=0
        Because it direct move between A and B are disallowed,We can first move T(n-1) from A to B,the move the largest
        disk to C(Move from A to B is disallowed),then move T(n-1)  from B to A,then move the largest disk in C to B,then
        remove the T(n-1) in A to B,so maximum moves is : T(n-1)+1+T(n-1)+1+T(n-1)
        T(0) = 0
        T(n) = 3T(n-1)+2
        Ok,Let's convert the recurrence  to a "closed form":
        We add 1 to both sides of equations:
        T(0) + 1 = 1;
        T(n) + 1 = 3T(n-1)+3
        Then we let U(n) = T(n)+1, we get :
        U(0) = 1;
        U(n) = 3(T(n-1)+1) = 3U(n-1))
        so the closed form is :  T(n) = 3^n -1