The number of tangents to the curve $xy - 3y + 2 = 0$ having slope 2 is:

The tangent line to the curve must have a slope of 2 at the point of tangency.

Given curve: $xy - 3y + 2 = 0$

Factoring out $y$:

$y(x - 3) = -2$

$y = \frac{-2}{x - 3}$

The curve is not defined at $x = 3$.

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Finding the derivative to determine the slope at any point:

$y = -2(x - 3)^{-1}$

$\frac{dy}{dx} = -2 \times (-1)(x - 3)^{-2}$

$\frac{dy}{dx} = \frac{2}{(x - 3)^2}$

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Setting the slope equal to 2:

$\frac{2}{(x - 3)^2} = 2$

$\frac{1}{(x - 3)^2} = 1$

$(x - 3)^2 = 1$

$x - 3 = \pm 1$

This gives two solutions:

$x - 3 = 1$ implies $x = 4$

$x - 3 = -1$ implies $x = 2$

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At $x = 4$:

$y = \frac{-2}{4 - 3} = -2$

Point: $(4, -2)$

At $x = 2$:

$y = \frac{-2}{2 - 3} = 2$

Point: $(2, 2)$

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There are 2 distinct points on the curve where the tangent line has slope 2.

Therefore, the number of tangents to the curve with slope 2 is $\boxed{2}$