1. As shown in Figure 7, let the equation of the line $AB$ be $y = kx + b$.
Substituting $x = \frac{3}{2}$ and $y = -\frac{7}{4}$, we get
$$b = -\frac{7}{4} - \frac{3}{2}k.$$
Therefore, $y = kx - \frac{3}{2}k - \frac{7}{4}$.
From the system of equations
$$\left\{\begin{array}{l} y = kx - \frac{3}{2}k - \frac{7}{4}, \\ y = x^2 - 4x + 3, \end{array}\right.$$
eliminating $y$ gives
$$x^2 - (4 + k)x + \left(\frac{3}{2}k + \frac{19}{4}\right) = 0.$$

Since the line $AB$ is tangent to the parabola, we have
$$\Delta = (4 + k)^2 - 4\left(\frac{3}{2}k + \frac{19}{4}\right) = 0.$$

Solving this, we get $k = 1$ or $k = -3$.
Given that $k > 0$, we have $k = 1$.
Therefore, the equation of the line $AB$ is $y = x - \frac{13}{4}$. It is easy to see that $A\left(\frac{13}{4}, 0\right)$ and $B\left(0, -\frac{13}{4}\right)$, so $OA = \frac{13}{4}$ and $OB = \frac{13}{4}$.
In the right triangle $\triangle AOB$, $AB = \sqrt{OA^2 + OB^2} = \frac{13\sqrt{2}}{4}$.
Let the distance from point $P$ to the line $AB$ be $d$. Connecting $PO$, $PA$, and $PB$, we have
$$\frac{1}{2}OA \cdot 1 + \frac{1}{2}OB \cdot 2 + \frac{1}{2}AB \cdot d = \frac{1}{2}OA \cdot OB.$$

Considering $P(2, -1)$, we get
$$d = \frac{OA \cdot OB - OA - 2OB}{AB} = \frac{\sqrt{2}}{8}.$$
Therefore, the distance from point $P$ to the line $AB$ is $\frac{\sqrt{2}}{8}$.