Equating the two given curveto find of intersection we have,

$3x^2+7x+4=x^3-3x^2-8x-4$

$\Rightarrow x^3-6x^2-15x-8=0$

$\Rightarrow {\left(x+1\right)}^2\left(x-8\right)=0$

Now putting $x=-1$ in the $y=3x^2+7x+4$ equation, we get

$y=3-7+4=0$,

Now putting $x=8$ in the equation $y=3x^2+7x+4$ we get $y=252$

So, the points are $\left(8,252\right)$ and $\left(-1,0\right)$

Now for equation of the tangent -

First we will find slope $\frac{dy}{dx}=6x+7$

So, at $x=-1 \text{or} 8$ we get $\frac{dy}{dx}=1$ or $55$

Now finding equation we get,

$y=mx+c$

$y=x+c \mathrm{or} y=55x+c$

Now given tangent passing through $\left(-1,0\right) \text{or} \left(8,252\right)$ so we get $c=1 \mathrm{or} c=188$

So, equation will be  $y=x+1$ or $y=55x+188$