China Mathematical Competition

$x^2-ax-4=0$ has two roots: $$x_1=\frac{a}{2}-\sqrt{4+\frac{a^2}{4}},x_2=\frac{a}{2}+\sqrt{4+\frac{a^2}{4}}$$ We have $B\subseteq A\iff x_1\ge -2$ and $x_2<4$. This means that $$\frac{a}{2}-\sqrt{4+\frac{a^2}{4}}\ge -2,\frac{a}{2}+\sqrt{4+\frac{a^2}{4}}<4$$ From the above we get $0\le a<3$.