I'm assuming you don't want any calculus involved here (too bad!), so let $\,(a,b)\,,\,(c,d)\,$ be the points on the parabola through which pass two tangent lines to it that are perpendicular:
$$\text{First tangent: we need to solve the system}\;\;\;\;y^2=4px\;\;,\;\;y-b=m(x-a)\Longrightarrow$$
$$(m(x-a)+b)^2=4px$$
$$\text{Second tangent: we need to solve the sytem}\;\;\;\;y^2=4px\;\;,\;\;y-d=-\frac{1}{m}(x-c)\Longrightarrow$$
$$\left(-\frac{1}{m}(x-c)+d\right)^2=4px$$
Of course, solving the above take into account that
$$b^2=4pa\;\;,\;\;d^2=4pc$$
since both points were chosen to be on the parabola.
Also, remember that two straight lines (none of which is horizontal/vertical) with slopes $\,m_1\,,\,m_2\,$ are perpendicular iff $\,m_1m_2=-1\,$, and this the reason we took the second tangent's slope to be $\,-1/m\,$
Yes, they are the same. In the textbook method, $2ydy=4pdx\rightarrow \frac{dy}{dx}=\frac{2p}{y}$. The slope $m=\frac{B}{A}=\frac{2p}{y}$. So $y=\frac{2p}{m}$.
Plugging this into the original equation $y^2=4px$ gives you $x=\frac{p}{m^2}$. Using the slope-intercept form of a line $y=mx+b$ would give you $b=\frac{p}{m}$.
Best Answer
Hint: remember that the equation of a parabola is $y = ax^2 + bx +c$ and $c$ is the intersection with the axis y (that you know in your exercise). In addition, you have to solve a system between your generic parabola and the tangent line and impose that they are only one point of intersection (putting $\Delta = 0$). The last condition is $- \frac{b}{2a} =\frac{1}{2}$. In this way, you'll find $a$, $b$ and $c$.
It's simpler if you use before the condition $c = -3$ and $- \frac{b}{2a} =\frac{1}{2}$ and then solve the system.