1. For all real numbers $p$ satisfying $0 \leqslant p \leqslant 4$, the inequality $x^{2}+p x>4 x+p-3$ always holds. Then the range of $x$ is ( ).
(A) $x>3$
(B) $-13$ or $x<-1$
-、1.D.
Transform the original inequality to get
$$
p(x-1)+\left(x^{2}-4 x+3\right)>0 \text {. }
$$

Let $f(p)=(x-1) p+\left(x^{2}-4 x+3\right)$. Then $f(p)$ is a linear function of $p$. From the original inequality, we have
$$
\left\{\begin{array} { l } 
{ f ( 0 ) > 0 , } \\
{ f ( 4 ) > 0 }
\end{array} \text { that is } \left\{\begin{array}{l}
x^{2}-4 x+3>0, \\
x^{2}-1>0 .
\end{array}\right.\right.
$$

Solving this, we get $x>3$ or $x<-1$.