Find the equations of these tangents and their point of intersection. I've done the first bit of the question, I just don't know how to do this part. 
a) Differentiate $$y=a(x-1)(x-5)$$ $$ y' = 2a(x-3)$$
b) Show that the tangents of the $x-$ intercepts $(1,0)$ and $(5,0)$ have opposite gradients: $y'(1)= -4a$ and $y'(5)= 4a$.
c) Find the equations of these tangents and their point of intersection.
Answer for c), according to the textbook, is $y=-4ax+ 4a$ and $y=4ax -20ax$
 A: Notice, $$y=a(x-1)(x-5)$$
a) $$\frac{dy}{dx}=a(x-1)+a(x-5)=2a(x-3)$$ 
b) Now, gradient of tangent at $(1, 0)$ $$m_1=2a(1-3)=-4a$$
Similarly, gradient of tangent at $(5, 0)$ $$m_1=2a(5-3)=4a$$ we find that the gradients of both the tangents are opposite in sign 
c) Equation of the tangent passing through $(1, 0)$ & having slope $m_1=-4a$ 
$$y-0=-4a(x-1)$$ $$\color{blue}{y=-4ax+4a}\tag 1$$
Equation of the tangent passing through $(5, 0)$ & having slope $m_1=4a$ 
$$y-0=4a(x-5)$$ $$\color{blue}{y=4ax-20a}\tag 2$$
Now, solving (1) & (2), we get point of intersection of the tangents as follows $$\color{blue}{\left(3, -8a\right)}$$
A: The other post is 100% correct but since you actually are only asking for the last part I thought I could do that part in some detail. 
We know by now that $y′=2a(x−3)$ and $y'(1) = -4a$,$y'(5)=4a$
c) Find the equations of these tangents and their point of intersection.
The formula for such a tangent line would be
$z-z_0=y'(x_0)(x-x_0)$
Basic idea of this formula is that you know the slope of the line should be the same as the derivative, and in a linear equation the slope is what is multiplied by the variable. In addition we need to shift this line which has the correct slope. The shifting is done by replacing $z \rightarrow z-z_0$ and $ x \rightarrow  x-x_0$ in a standard linear equation of the form $z=y'(x_0)x$
More info at http://www.math.brown.edu/utra/tangentline.html
If you do not like this formula,for whatever reason, you can alternatively do as follows:
We want a linear function so it should be on the form of 
$z=kx+b$
We know the slope is $k=y'(x_0)$, so all we need to figure out now is what the constant $b$ should be. Well we have a point clearly defined on the slope in the form of $(x_0,z_0)$ so we can simply solve the equation with $z=z_0$ and $x=x_0$, solving for b. For instance for the tangent passing through $(1,0)$ we get
$z=kx+b=0=-4a\cdot1+b$
solving it gives $b=4a$. 
so $z=-4ax+4a$
A good tutorial on this way can be found at https://www.khanacademy.org/math/differential-calculus/taking-derivatives/product_rule/v/equation-of-a-tangent-line
I will solve the problem using the first of the two options. 
When you have this equation, or the understanding of it you can simply use it on your problem the way Harish has done in his post. 
So for the first point $(1,0)$ with y'(1)=-4a
$z-z_0=y'(x_0)(x-x_0)\leftrightarrow z-0=-4a(x-1)$
And for the second point $(5,0)$ with y'(5)=4a
$z-z_0=y'(x_0)(x-x_0)\leftrightarrow z-0=4a(x-5)$
Now we have the two linear equations we were after and can solve by asking which x will yield the same height(z) for both equations. 
$4a(x-5)=-4a(x-1)$
which we can solve to yield $x=3$, put it into either linear equation and get $z=-8a$.
So then the intersection is $(3,-8a).
Feel free to comment, this is my first post done. 
