已知 $f(x)=x^2-ax+3$，其中 $a$ 为常数，

1. Set up the inequality and rearrange it into standard quadratic form

$x^2 - ax + 3 > x - 1$
$x^2 - ax - x + 4 > 0$
$x^2 - (a + 1)x + 4 > 0$

2. Apply the discriminant condition for a quadratic to be always positive

For the quadratic to be always positive, the discriminant must be less than zero (since the coefficient of $x^2$ is $1 > 0$).
$\Delta < 0$
$(-(a + 1))^2 - 4(1)(4) < 0$

3. Simplify the discriminant inequality

$(a + 1)^2 - 16 < 0$
$(a + 1)^2 < 16$

4. Solve the quadratic inequality for $a$

$-4 < a + 1 < 4$
$-5 < a < 3$

1. Form a quadratic equation by equating the line and the curve

$x^2 - ax + 3 = a + 4$
$x^2 - ax - a - 1 = 0$

2. Apply the condition for a line to be a tangent to a curve

For the line to be a tangent, the discriminant must be zero ($b^2 - 4ac = 0$).
$(-a)^2 - 4(1)(-a - 1) = 0$
$a^2 + 4a + 4 = 0$

3. Solve the resulting quadratic equation for $a$

$(a + 2)^2 = 0$
$a = -2$