( 1)

The first small problem, an+ 1=an holds, then 5-6/an=an, (2)

And because a 1=a, then an=a, substitute (2), and get a=2 or 3.

The second question, an+1>; An, that is, 5-6/an > Solving an < 0,2 <; An<3.

If we imagine one; 5, then A3; 0

So let's assume that 2

So 2 < a<3. Next, it can be proved by mathematical induction, so I won't write it in detail.

Thirdly, it can be seen that bn should be opposite to an, that is, bn=5-6/b(n+ 1), so that b(n+ 1)=6/(5-bn) and b 1= 1.

What we still need to prove is BN.

So let's assume that bn is b(n+ 1)=6/(5-bn) and b 1= 1, which satisfies two conditions.