Step 1: Simplify the curve.
The curve $y^2-2x^3-4y+8=0$ becomes, after completing the square in $y$, $(y-2)^2=2x^3-4$. So $(y-2)^2=2(x^3-2)$.
Step 2: Parametrise a point of contact.
Let the point of tangency be $(a,b)$ on the curve, so $(b-2)^2=2(a^3-2)$. We seek tangents from the external point $(1,2)$.
Step 3: Find the slope by implicit differentiation.
Differentiating $y^2-2x^3-4y+8=0$ gives $2y\,y'-6x^2-4y'=0$, so $y'=\frac{6x^2}{2y-4}=\frac{3x^2}{y-2}$. At $(a,b)$ the slope is $\frac{3a^2}{b-2}$.
Step 4: Impose that the tangent passes through $(1,2)$.
The slope from $(a,b)$ to $(1,2)$ must equal the tangent slope: $\frac{b-2}{a-1}=\frac{3a^2}{b-2}$, giving $(b-2)^2=3a^2(a-1)$.
Step 5: Combine with the curve relation.
Since $(b-2)^2=2a^3-4$, equate: $2a^3-4=3a^3-3a^2$, which rearranges to $a^3-3a^2+4=0$. Factoring, $(a+1)(a-2)^2=0$, so $a=-1$ or $a=2$ (double).
Step 6: Keep only real points of contact.
For $a=-1$: $(b-2)^2=2(-1)-4=-6<0$, impossible. For $a=2$: $(b-2)^2=2(8)-4=12$, which gives real $b$, and the repeated root corresponds to a single tangent line. Hence exactly one tangent passes through $(1,2)$. \[ \boxed{1} \]