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 $$g(t)=t^2+at+a-\frac{3}{a}.$$ The sufficient and necessary condition for $f(x)\le 0,\forall x\in \mathbf{R}$ is $$\{\begin{array}{l}g(-1)=1-\frac{3}{a}\le 0,\\ g(1)=1+2a-\frac{3}{a}\le 0.\end{array}$$ 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 $[2,3]$.