I think so.
Let's start with electron-phonon scattering.
Low temperature
You know that phonons are goldstone bosons with a dispersion of $\omega = ck$ at low momentum. Now imagine you're at low temperature. What will be the density of phonons in your material? Well, number of phonons per unit volume, by dimensional analysis, should go like $1/\lambda^3$ where $\lambda = c/T$ is the phonon de Broglie wavelength. Assuming the rate of scattering is proportional to the density of scatters (phonons) you find that the electron scattering time is proportional to $T^3$. However, this isn't quite what you want, because you want to know how much the current is degraded by a phonon scattering event.
It turns out that low temperature phonons are bad at this because low temperatures phonons have very small typical wavevectors of order $1/\lambda \sim T/c$. Hence when they impart a little kick to electrons near the Fermi surface with momentum $k \sim k_F$ they barely change the electron's momentum and barely degrade the current. In the jargon, we have the "scattering time" $\tau$ and the "transport time" $\tau_{tr}$ which are schematically related as $\frac{1}{\tau_{tr}} \sim (1-\cos{\theta}) \frac{1}{\tau} $ where $\theta$ is the scattering angle. (The actual computation proceeds by summing this sort of formula over final states, but we can just use a typical value.) The angle dependent factor is just saying that the actual loss of current only comes when the final state is significantly different from the initial state (try drawing a picture of the currents and computing the change along the original direction).
To complete the picture we need to understand the physics of this extra angle dependent factor. The angle $\theta$ is small and is typically of order $1/(\lambda k_F)$ e.g. 795
if the phonon adds a little momentum kick at right angles to the electrons momentum. Expanding $1-\cos{\theta}$ you get two extra powers of $T$ for a inverse transport time of order $T^5$. Plug this into the <PERSON> formula to get your answer (I'm assuming you know this formula; if not, rest assured that its also physically sensible. I can briefly describe it later if you're interested).
High temperature
This one is easier. At high temperatures the phonons behave like classical springs. Equipartition of energy in a classical oscillator tells you that that the energy and hence the number of phonons now goes like $T$ (the units are made up by the Debye frequency or something like it). Furthermore, at these high temperatures the transport time is not so different from the scattering time since phonons are much better at imparting large momentum. Hence the resistivity now goes like $T$.
If this was intuitive to you, I can also discuss the situation with electron-electron interactions. If not, let me know and I can try to give a more appropriate response. 795