How does the Freezing Point fall in a solution?
Focus more on free energies rather than on vapor pressures (which derive, ultimately, from free energies after all).  For a mixture of B (solute) in A (solvent), the entropy of mixing is $RT(x_A ln(x_A) + x_B ln(x_B))$, and the enthalpy of mixing will go approximately as $x_A x_B \Omega$, with $\Omega$ as a measure of the interaction of A and B. The entropy term will always result in a reduction of free energy at small $x_B$ regardless of the sign of $\Omega$, but in the case of salt in water $\Omega$ is negative, driving further solubility. 

Now, about those temperatures of phase transitions (which as you should recall occur when the free energies of the phases are equal, more fundamental than vapor pressures).  On the boiling end, water with salt in it has a lower free energy than water without salt - the boiling point of the salt water has to be higher than for pure water.  The presence of the solute makes it happy to stay liquid.

So, why isn't the freezing point raised in an analogous way? For pure A, of course, the freezing point remains that of pure water.  The entropy of mixing is equivalent for solid and liquid in this case, and it is not clear how different the excess enthalpy is going to be - so why does liquid continue to be stable?  Well, it only does while it has a higher concentration of salt in it than is in the solid.  It forms a classic eutectic point on the binary phase diagram (see, for example http://antoine.frostburg.edu/chem/senese/101/solutions/images/saltwater-phase-diagram.gif).  The only requirement of melting point depression is that water at a high concentration of salt has a lower free energy than the solid at a lower concentration of salt.  Given the outline of free energy above, this is likely to hold for some range of temperatures. 

Again, remember that what is being lowered is the freezing point of water with salt in it - what starts freezing out is ice with a lower salt content in it.