Why does NaBH4 reduce double bonds conjugated to carbonyl groups, while LiAlH4 does not?
Neither $\ce{LiAlH_4}$ nor $\ce{NaBH4}$ are able to reduce an isolated $\ce{C=C}$ bond. But if you have an enal (a conjugated aldehyde) it can react (as an electrophile) either at the β-carbon or at the carbonyle group's carbon. According to the HSAB Principle the β-carbon is a “soft” center and would react preferably with “soft” nucleophiles while the carbonyle carbon is a “hard” center and prefers to react with “hard” nucleophiles.
Now, $\ce{NaBH4}$ is a rather soft nucleophile and thus it reacts with an enal at the β-carbon. After this reaction the $\ce{C=C}$ bond is gone. But what is left behind is still a very reactive simple aldehyde that gets reduced to the alcohol by $\ce{NaBH4}$ in a second step.
$\ce{LiAlH4}$ on the other hand is a rather hard nucleophile and thus reacts with an enal at the carbonyl carbon. After this reaction there is just an isolated double bond left that can't be reduced by $\ce{LiAlH4}$ in a second step.