(1) for the inequality on the left, there are:

A+1+1-2 (sinx) 2+sinx-a2 ≤ 0 means:

2(sinx)^2-sinx+a^2-a-2≥0

Let t=sinx, then t∈[- 1, 1]

Let f(t) = 2t 2-t+A2-A-2, so f (t) is always greater than or equal to 0 when t∈[- 1].

Because the opening of f(t) is upward, the symmetry axis is t= 1/4.

Therefore, if f(t) is always greater than or equal to 0 at t∈[- 1], it needs to satisfy:

δ = 1-4× 2 (a 2-a-2) ≤ 0, and the solution is:

A≥(2+√38)/4 or a≤(2-√38)/4.

② For the inequality on the right:

A 2-sinx ≤ 3, that is, A 2-3 ≤ sinx holds.

So: A 2-3 ≤- 1 gives:

-√2≤a≤√2

From ① ②: -√2≤a≤(2-√38)/4