(2) Let $A=[-2,4), B=\left\{x \mid x^{2}-a x-4 \leqslant 0\right\}$, if $B \subseteq A$, then the range of real number $a$ is ( ).
(A) $[-1,2)$
(B) $[-1,2]$
(C) $[0,3]$
(D) $[0,3)$
(2. Since $x^{2}-a x-4=0$ has two real roots
$$
x_{1}=\frac{a}{2}-\sqrt{4+\frac{a^{2}}{4}}, x_{2}=\frac{a}{2}+\sqrt{4+\frac{a^{2}}{4}},
$$

thus $B \subseteq A$ is equivalent to $x_{1} \geqslant-2$ and $x_{2}<4$, that is
$$
\frac{a}{2}-\sqrt{4+\frac{a^{2}}{4}} \geqslant-2
$$

and
$$
\frac{a}{2}+\sqrt{4+\frac{a^{2}}{4}}<4,
$$

solving these yields $0 \leqslant a<3$. Therefore, the answer is $\mathrm{D}$.