I only know about why probability of finding e- @ nucleus is not zero in s orbitals. Out teacher told us that when we find the probality to find the electron in 3D space, we make a spherical shell of radius r and width dr and then find the probability there.
From here P comes out to be (psi)^2 x dV, which comes out to be (psi)^2 4(pi)r^2.dr so when r = 0, the probability comes out to be zero in nucleus (i.e. r = 0).
It's the probability of finding an electron there at r = 0 is zero and not the wave function. @Supreeta_Sen
As far as wave function is concerned, when we normalize Schrodinger wave functions of s-orbitals about the (/theta) & (/phi) spherical co-oordinates (only applicable for H) the (/psi)(r) have inverse value of r. So, when r = 0, the wave function becomes infinitesimal.
Not sure about p-orbitals tho. Sir taught us about s-orbitals only. I will ask him tommorow, for sure.
P.S. MathJAX is broken (or I donno how to use it.)