China Mathematical Competition

(1) We have $f(x) = \sin^2 x + a \sin x + a - \frac{3}{a}$. Let $t = \sin x$ ($-1 \le t \le 1$). Then
$$g(t) = t^2 + a t + a - \frac{3}{a}.$$
The sufficient and necessary condition for $f(x) \le 0, \forall x \in \mathbf{R}$ is
$$\begin{cases} g(-1) = 1 - \frac{3}{a} \le 0, \\ g(1) = 1 + 2a - \frac{3}{a} \le 0. \end{cases}$$
Therefore, we obtain the range of $a$ is $(0, 1]$.

(2)
As $a \ge 2$, then $-\frac{a}{2} \le -1$. We have
$$g(t)_{\min} = g(-1) = 1 - \frac{3}{a}.$$
Then $f(x)_{\min} = 1 - \frac{3}{a}$. Therefore, the sufficient and necessary condition for $f(x) \le 0, \exists x \in \mathbf{R}$ is $1 - \frac{3}{a} \le 0$, or $0 < a \le 3$.
Finally, we obtain that the range of $a$ is $[2, 3]$.