1) I know what it means for a finite set of points in a projective space to be in general position, but what does it mean for two quadrics in $\mathbb{P}^3(\mathbb{C})$ to be in general position?
2) Is it true that the intersection of two smooth quadrics in $\mathbb{P}^3(\mathbb{C})$ is also smooth? If not, what additional assumption does one need to ensure the smoothness of the intersection?
Please keep in mind that my background in both topology and algebraic geometry is virtually non-existent, I would be extremely grateful if you could answer assuming only a basic knowldege of projective geometry.
You need the two smooth quadrics $Q_1$ and $Q_2$ to intersect transversely. This means that at all points of $Q_1\cap Q_2$ the two surfaces will have distinct tangent planes. As usual in multivariable calculus, you can check this by making sure that the gradients (or differentials) of the quadratic defining functions are linearly independent at all such points. This does in fact happen "generally."
Indeed, it follows from Bertini's Theorem that the general element of the linear system $|2H|$ on $Q_1$ will be smooth. (You can also deduce this from standard transversality theory. Intuitively, the nonsingular matrices of rank $<2$ in the set of complex $2\times 4$ matrices is a submanifold of codimension $3$, and so it's $3$ (complex) conditions on the space of functions $(f_1,f_2)\colon\Bbb C^4\to \Bbb C^2$ for the derivative to land in that submanifold at some point.)