From the given equation of ellipse, we have

$$a = 4,b = 3,e = \sqrt {1 - {9 \over {16}}} $$

$$ \Rightarrow e = {{\sqrt 7 } \over 4}$$

Now, radius of this circle $$ = {a^2} = 16$$

$$ \Rightarrow Focii = \left( { \pm \sqrt 7 ,0} \right)$$

Now equation of circle is

$${\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = 16$$

$${x^2} + y{}^2 - 6y - 7 = 0$$

Equation of circle is $${\left( {x - 1} \right)^2} + {y^2} = 1$$

$$ \Rightarrow $$ radius $$=1$$ and diameter $$=2$$

$$\therefore$$ Length of semi-minor axis is $$2.$$

Equation of circle is $${x^2} + {\left( {y - 2} \right)^2} = 4 = {\left( 2 \right)^2}$$

$$ \Rightarrow $$ radius $$=2$$ and diameter $$=4$$

$$\therefore$$ Length of semi major axis is $$4$$

We know, equation of ellipse is given by

$${{{x^2}} \over {\left( {Major\,\,\,axi{s^{\,\,2}}} \right)}} + {{{y^2}} \over {\left( {Minor\,\,\,axi{s^{\,\,2}}} \right)}} = 1$$

$$ \Rightarrow {{{x^2}} \over {{{\left( 4 \right)}^2}}} + {{{y^2}} \over {{{\left( 2 \right)}^2}}} = 1$$

$$ \Rightarrow {{{x^2}} \over {16}} + {{{y^2}} \over 4} = 1$$

$$ \Rightarrow {x^2} + 4{y^2} = 16$$

Given equation of ellipse is $$2{x^2} + {y^2} = 4$$

$$ \Rightarrow {{2{x^2}} \over 4} + {{{y^2}} \over 4} = 1 \Rightarrow {{{x_2}} \over 2} + {{{y^2}} \over 4} = 1$$

Equation of tangent to the ellipse $${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$$ is

$$y = mx \pm \sqrt {2{m^2} + 4} \,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

( as equation of tangent to the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

is $$y=mx+c$$ where $$c = \pm \sqrt {{a^2}{m^2} + {b^2}} $$ )

Now, Equation of tangent to the parabola

$${y^2} = 16\sqrt 3 x$$ is $$y = mx + {{4\sqrt 3 } \over m}\,\,\,\,\,\,\,\,...\left( 2 \right)$$

( as equation of tangent to the parabola

$${y^2} = 4ax$$ is $$y = mx + {a \over m}$$ )

On comparing $$(1)$$ and $$(2),$$ we get

$${{4\sqrt 3 } \over m} = \pm \sqrt {2{m^2} + 4} $$

Squaring on both the sides, we get

$$16\left( 3 \right) = \left( {2{m^2} + 4} \right){m^2}$$

$$ \Rightarrow 48 = {m^2}\left( {2{m^2} + 4} \right) \Rightarrow 2{m^4} + 4{m^2} - 48 = 0$$

$$ \Rightarrow {m^4} + 2{m^2} - 24 = 0 \Rightarrow \left( {{m^2} + 6} \right)\left( {{m^2} - 4} \right) = 0$$

$$ \Rightarrow {m^2} = 4$$ ( as $${m^2} \ne - 6$$ ) $$ \Rightarrow m = \pm 2$$

$$ \Rightarrow $$ Equation of common tangents are $$y = \pm 2x \pm 2\sqrt 3 $$

Thus, statement - $$1$$ is true.

Statement - $$2$$ is obviously true.

Let the ellipse be $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

It press through $$(-3, 1)$$ so $${9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)$$

Also, $${b^2} = {a^2}\left( {1 - 2/5} \right)$$

$$ \Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$$

So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$