Statement-1: The line $$x - 2y = 2$$ meets the parabola, $$y^2 + 2x = 0$$ only at the point (-2, -2).
Statement-2: The line $$y = mx - \frac{1}{2m}$$ ($$m \neq 0$$) is tangent to the parabola, $$y^2 = -2x$$ at the point $$\left(-\frac{1}{2m^2}, -\frac{1}{m}\right)$$

We have the parabola $$y^{2}+2x=0$$ which can be rewritten as $$x=-\dfrac{y^{2}}{2}$$. The given straight line is $$x-2y=2$$, that is $$x=2y+2$$.

To find their points of intersection we substitute the value of $$x$$ from the line into the parabola:

$$-\dfrac{y^{2}}{2}=2y+2$$

Multiplying both sides by $$2$$ to clear the denominator,

$$-y^{2}=4y+4$$

Now bring every term to the left side:

$$y^{2}+4y+4=0$$

Notice that the left-hand side is a perfect square:

$$(y+2)^{2}=0$$

So,

$$y=-2$$

Substituting this value of $$y$$ back in $$x=2y+2$$,

$$x=2(-2)+2=-4+2=-2$$

Hence, the only common point of the line and the parabola is $$(-2,-2)$$, and because the quadratic equation in $$y$$ gives a repeated root, the line merely touches (is tangent to) the parabola there. Therefore Statement-1 is true.

Now we examine Statement-2. First recall the slope-form of the tangent to the standard parabola $$y^{2}=4ax$$. It is stated in textbooks as

$$y=mx+\dfrac{a}{m},\qquad m\neq 0$$

where $$m$$ is the slope and $$\left(\dfrac{a}{m^{2}},\,\dfrac{2a}{m}\right)$$ is the corresponding point of contact.

For the parabola in the present question, $$y^{2}=-2x$$. We equate this to the standard form $$y^{2}=4ax$$ to find $$a$$:

$$4a=-2\;\;\Longrightarrow\;\;a=-\dfrac12$$

Using the above formula for the tangent with this value of $$a$$, we get

$$y=mx+\dfrac{a}{m}=mx+\dfrac{-\dfrac12}{m}=mx-\dfrac1{2m},\qquad m\neq 0$$

which is exactly the line quoted in Statement-2.

Further, the point of contact should be

$$\left(\dfrac{a}{m^{2}},\,\dfrac{2a}{m}\right)=\left(\dfrac{-\dfrac12}{m^{2}},\,\dfrac{2\!\left(-\dfrac12\right)}{m}\right)=\left(-\dfrac1{2m^{2}},\,-\dfrac1{m}\right)$$

This is exactly the point written in Statement-2. Hence Statement-2 is also true.

Finally, observe that by putting $$m=\dfrac12$$ in the general tangent $$y=mx-\dfrac1{2m}$$ we obtain

$$y=\dfrac12x-\dfrac1{2\left(\tfrac12\right)}=\dfrac12x-1$$

which, after multiplying by $$2$$, gives $$x-2y=2$$ - the very line of Statement-1. The corresponding point of contact becomes

$$\left(-\dfrac1{2\left(\tfrac12\right)^{2}},\,-\dfrac1{\tfrac12}\right)=\left(-2,\,-2\right)$$

Thus Statement-2 supplies the general fact from which Statement-1 follows as a direct special case. Therefore Statement-2 is indeed a correct explanation for Statement-1.

Hence, the correct answer is Option B.