(Given: radial function = r^αe^−βr and ϕ=e^−3iϕ)
From Schrodinger equation for H-atom the azimuthal angular part (Φ) is:
Φ(ϕ)=e^(±imϕ)
Where,
Φ(ϕ) represents the azimuthal angular part of the wave function.
(1/√(2π)) is the normalization constant.
e^(imϕ) is the complex exponential function.
m is the magnetic quantum number, where m can be positive or negative including zero.
ϕ is the azimuthal angle
Now, let's analyze the options:
(1) n > 4, l > 3, m = 3
Here, n>4 and l>3, assume n = 5 and l = 5 for m = 3. The combination of values of n, l, m is typically not found in a stable atomic system because combining a high 'n' with a high 'l' suggests an electron in an extremely high-energy, complex orbital with many radial and angular nodes. Therefore, this option is not correct.
(2) n = 4, l = 3, m = -3
Here, n = 4 and l = 3, indicates that the electron is in the fourth energy level or shell in an f orbital. m = -3, which means the electron is in an f orbital with a specific orientation. Therefore, this option is correct.
(3) n = 4, l > 3, m = 3
Here, n = 4 and l>3 shows combination of relatively high energy state (n = 4) with a complex orbital type (l > 3) that has complex angular distributions (e.g., f orbitals). Such a combination is not observed in stable atomic systems. Rather, electrons tend to occupy lower energy levels with simpler orbitals before filling higher energy levels with more complex orbitals. This makes the combination of n = 4 and l > 3 unusual for stable atomic configurations. Therefore, this option is not correct. 
(4) n > 4, l = 3, m = -3
Here, n>4 and l = 3, also combines higher energy state with complex angular distribution (eg. f orbital) which makes it unusual for stable atomic system. Therefore, this option is incorrect.
Therefore, the radial and angular function of hydrogenic orbital corresponds to n = 4, l = 3, m = -3.