From the point $(1,2),$ tangents are drawn to the curve  $y^2-2x^3-4y+8=0.$ Then

Curve is  ${(y-2)}^2=2x^3-4\mathrm{.....}\left(1\right)$
 $\frac{dy}{dx}=\frac{3x^2}{y-2}$
Equation of tangent at  $(x_1,y_1) \text{\hspace{0.17em}is\hspace{0.17em}}y-y_1=\frac{3x_1^2}{y_1-2}(x-x_1)$
As passes through $(1,2)$
$$\begin{gathered} \Rightarrow 2-y_1=\frac{3x_1^2}{y_1-2}(1-x_1) \\ \Rightarrow {(y_1-2)}^2=3x_1^2(x_1-1) \\ \Rightarrow 2x_1^2-4=3x_1^3-3x_1^2 \\ [as {(y_1-2)}^2=2x_1^3-4] \\ \Rightarrow x_1^3-3x_1^2+4=0 \\ \Rightarrow (x_1+1){(x_1-2)}^2=0 \\ \Rightarrow x_1=-1or2 \end{gathered}$$

From equation (1) it is clear that  $x_1\ne -1$
Thus  $z_1=2\Rightarrow y_1=2\pm 2\sqrt{3}$
Points of contact are $(2,2+2\sqrt{3})$ and  $(2,2-2\sqrt{3})$
$\therefore$  sum of x-coordinates of points of contact is 4
$\therefore$ sum of y-coordinates of points of contact is 4