Let $$P\left( {at_1^2,\,2a{t_1}} \right)$$   and $$Q\left( {at_2^2,\,2a{t_2}} \right)$$   be two points on the parabola $${y^2} = 4ax.$$   The tangents at $$P$$ and $$Q$$ intersect at $$T\left[ {a{t_1}{t_2},\,a\left( {{t_1} + {t_2}} \right)} \right].$$
$$\eqalign{ & {\text{Now, }}SP = \sqrt {{{\left( {at_1^2 - a} \right)}^2} + 2{{\left( {a{t_1} - 0} \right)}^2}} = a\left( {t_1^2 + 1} \right)\,; \cr & SQ = a\left( {t_2^2 + 1} \right) \cr & {\text{and }}ST = \sqrt {{{\left( {a{t_1}{t_2} - a} \right)}^2} + {{\left[ {a\left( {{t_1} + {t_2}} \right) - 0} \right]}^2}} \cr & \Rightarrow ST = a\sqrt {\left( {1 + t_1^2} \right)\left( {1 + t_2^2} \right)} \cr & \Rightarrow S{T^2} = {a^2}\left( {1 + t_1^2} \right)\left( {1 + t_2^2} \right) \cr & \Rightarrow S{T^2} = SP.SQ \cr} $$
Hence, $$SP,\,ST,\,SQ$$    are in G.P.

Any tangent to parabola $${y^2} = 8x$$  is $$y = mx + \frac{2}{m}......\left( {\text{i}} \right)$$
It touches the circle $${x^2} + {y^2} - 12x + 4 = 0,$$     if the length of perpendicular from the centre $$\left( {6,\,0} \right)$$  is equal to radius $$\sqrt {32} .$$
$$\eqalign{ & \therefore \,\frac{{6m + \frac{2}{m}}}{{\sqrt {{m^2} + 1} }} = \pm \sqrt {32} \cr & \Rightarrow {\left( {3m + \frac{1}{m}} \right)^2} = 8\left( {{m^2} + 1} \right) \cr & \Rightarrow {\left( {3{m^2} + 1} \right)^2} = 8\left( {{m^4} + {m^2}} \right) \cr & \Rightarrow {m^4} - 2{m^2} + 1 = 0 \cr & \Rightarrow m = \pm 1 \cr} $$
Hence, the required tangents are $$y = x + 2$$   and $$y = - x - 2.$$

$$y = mx + \frac{1}{m}$$
Above tangent passes through (1, 4)
$$ \Rightarrow 4 = m + \frac{1}{m}\,\,\, \Rightarrow {m^2} - 4m + 1 = 0$$
Now angle between the lines is given by
$$\eqalign{ & \tan \,\theta = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| = \frac{{\sqrt {{{\left( {{m_1} + {m_2}} \right)}^2} - 4{m_1}{m_2}} }}{{1 + {m_1}{m_2}}} \cr & = \frac{{\sqrt {16 - 4} }}{{1 + 1}} = \sqrt 3 \cr & \Rightarrow \theta = \frac{\pi }{3} \cr} $$