Your vf for the bullet is correct. As physixguru says, you do not need conservation of kinetic energy to solve the other two parts. Part (c) probably means the kinetic energy gained by the block, or else (b) and (c) contradict each other.
If you want to assume that kinetic energy is conserved (i.e. the collision between the bullet and the block is elastic), you will find that v0 does not cancel:
[tex]\text{From part (a), }v_{\text{block}} = \frac{v_0}{6}[/tex]
[tex]\text{From the assumption that kinetic energy is conserved, }[/tex]
[tex]mv_0 &=& m\cdot\left(\frac{v_0}{3}\right)^2 + 4m\cdot\left(\frac{v_0}{6}\right)^2 = \frac{mv_0^2}{9} + \frac{4mv_0^2}{36} = \frac{2mv_0^2}{9} \implies v_0^2 - \frac{9v_0}{2} = v_0\left(v_0 - \frac{9}{2}\right) = 0 \implies v_0 = 0 \text{ or } \frac{9}{2}[/tex]
So if the given conditions are met, kinetic energy is conserved when the initial velocity of the bullet is either 0 (it isn't fired) or 9/2.
However, it is unrealistic that kinetic energy is conserved (i.e. the collision is elastic), because the impact of the bullet against the wood and the friction on the bullet when it travels through the wood will generate heat, so some of the initial kinetic energy will be lost (i.e. the collision is inelastic).
Therefore, you really cannot find out what the initial velocity v0 is. For the given block-bullet system, there may be a particular v0 which will cause the exit speed of the bullet to be v0/3, but that speed is impossible to calculate and depends, among other factors, on the length of the block and the materials used.