To the last question, traditionally if one does not qualify the coefficients directly or from context, rational or integer coefficients are assumed, so the polynomial is a real function.
The calculations for the case with complex roots should be the same as with the real roots, just take $b,c$ complex instead of real, with $c=\bar b$.
$f(x)=(x-a)(x-b)(x-c)$, $f'(x)=(x-a)(x-b)+(x-a)(x-c)+(x-b)(x-c)$, the tangent at $x_0=\frac12(a+b)$ is
$$
t(x)=f(x_0)+(x-x_0)f'(x_0)=(x_0-a)(x_0-b)(x_0-c)+(x-x_0)(x_0-a)(x_0-b)
=(x-c)(x_0-a)(x_0-b)
$$
so it has a root at $x=c$.
However, you would get the third root more quickly by using that $-(a+b+c)$ is the quadratic coefficient of the polynomial.
Equations of tangents are $y=m_1x+\frac{a}{m_1}$ and $y=m_2x+\frac{a}{m_2}$
I think this is correct.
and equations of normal to the parabola $x^2=4by$ is $y=m_1x-2bm_1-bm^3_1$ and $y=m_2x-2bm_2-bm^3_2$.
I don't think this is correct.
Let $(s,t)$ be a point on $x^2=4by$, i.e. $s^2=4bt$. Since $2x=4by'\Rightarrow y'=\frac{x}{2b}$, the equation of normal to the parabola at $(s,t)$ will be $y-t=-\frac{2b}{s}(x-s)\iff y=-\frac{2b}{s}x+2b+t$. Let $k=-\frac{2b}{s}$. Then, since $s=-\frac{2b}{k}$, the equation of the normal can be written as
$$y=kx+2b+\frac{s^2}{4b}=kx+2b+\frac{1}{4b}\left(-\frac{2b}{k}\right)^2=kx+2b+\frac{b}{k^2}.$$
So, in our case, we have
$$y=m_1x+2b+\frac{b}{m_1^2}\qquad\text{and}\qquad y=m_2x+2b+\frac{b}{m_2^2}.$$
Thus, since $y=m_1x+\frac{a}{m_1}$ and $y=m_1x+2b+\frac{b}{m_1^2}$ are the same lines,
$$\frac{a}{m_1}=2b+\frac{b}{m_1^2}\iff 2bm_1^2-am_1+b=0.$$
Now, the discriminant is positive, so $(-a)^2-4\cdot 2b\cdot b\gt 0$, i.e. $a^2\gt 8b^2$.
As harry pointed out in the comments, one has to consider the case where one of the tangents is the $y$-axis.
Two tangents drawn to the parabola $y^2=4ax$ from $(0,k)$ where $k\not=0$ are $x=0$ and $y=\frac akx+k$. It follows from $u^2=4bv,\frac ak=-\frac{2a}{u}$ and $k=2b+v$ that $bk^2-a^2k+2a^2b=0$. So, one has to have $a^2\geqslant 8b^2$. If $a^2=8b^2$, then two tangents drawn to the parabola $y^2=4ax$ from a point $(0,4b)$ are $x=0$ and $ax-4by+16b^2=0$ (at $(2a,8b)$) which are normals to the parabola $x^2=4by$ (at $(-a,2b)$).
In conclusion, the answer is $a^2\geqslant 8b^2$.
Best Answer
The quoted proof appears to prove the opposite implication i.e. if the three normals are real then $\,h \gt 2a\,$, and also seems to assume that $\,a \gt 0\,$.
The statement that the proof actually proves is that when the three normals are real then the relation $h>2a$ is satisfied.
$m_1^2+m_2^2+m_3^2 \ge 0$ is a necessary condition for the three roots to be real, but it is not a sufficient condition, for example $\,(-2)^2 + (1+i)^2 + (1-i)^2 = 4\ge 0\,$ where $\,-2, 1+i, 1-i\,$ are the roots of the depressed cubic $\,m^3-2m+4\,$.
However, the proof only requires the forward implication three real roots $\,\implies m_1^2+m_2^2+m_3^2 \ge 0\,$, so being a necessary condition is enough for the purpose of the proof.