1. If the graph of the function with respect to $x$
$$
y=(a-3) x^{2}-(4 a-1) x+4 a
$$

intersects the coordinate axes at two points, then the value of $a$ is
$$
\text { II. 1.3, 0 or }-\frac{1}{40} \text {. }
$$

When $a-3=0$, i.e., $a=3$, the original function becomes $y=-11 x + 12$, and its graph intersects the coordinate axes at two points.

When $a-3 \neq 0$, i.e., $a \neq 3$, the original function is a quadratic function, which must intersect the $y$-axis at one point. If this intersection point is not the origin, it can only intersect the $x$-axis at one point. Therefore,
$$
\Delta=(4 a-1)^{2}-16 a(a-3)=0 .
$$

Solving for $a$ gives $a=-\frac{1}{40}$. If this intersection point is the origin, $a=0$, and it must intersect the $x$-axis at two distinct points.
In summary, $a=3, 0$ or $-\frac{1}{40}$.