Is Dropbox's hashing method cryptographically secure? Dropbox uses CRC32 and md5 to create a checksum of each 4 mb "block" as part of the their compression/file system.

I suspect that Dropbox uses an additional cryptographic hash to check if there's a collision, but let's assume that doesn't happen.

Is using CRC32 and md5 in combination cryptographically secure?
Disclaimer: I have no first-hand knowledge of what hash (or MAC, or whatever method) DropBox uses for de-duplication; about if it is enough to know that (and its key, for a MAC) in order to download something from DropBox; and I see slightly diverging opinions about these points.
If we consider the problem of finding a collision, the 160-bit hash defined by
$$H(M)=\text{CRC32}(M)||\text{MD5}(M)$$
is NOT cryptographically secure, for it is only marginally stronger than $\text{MD5}$ is.
$\text{MD5}$ no longer provides a good protection against collisions: with this Fast Collision Attack on $\text{MD5}$ we can find a collision (for messages at least 128 byte) with cost about $2^{18}$ compression functions (finding collisions for $\text{CRC32}$ is totally trivial). Finding a collision for $H$ is harder than for $\text{MD5}$, but by a dumb method (finding $\text{MD5}$ collisions until one is also a collision for $\text{CRC32}$) only like $2^{32}$ times harder, and $2^{50}$ compression functions is non-trivial but feasible. Update: And as pointed by poncho we can (for messages at least 4 kiB) make that only like $32$ times harder, or about $2^{23}$ compression functions, which is nothing. As the saying attributed to the NSA goes, attacks only get better, they never get worse.

If we consider the problem of finding a second-preimage, $\text{MD5}$ remains impractical to attack as far a we know, and $H$ is at least as strong as $\text{MD5}$ is.
If we consider the problem of finding a first-preimage, $H$ can't be more than $2^{32}$ times easier to attack than $\text{MD5}$ is, likely that can only be approached for short messages where brute-force is the best attack, and for long messages likely this is (at least) about as hard as for $\text{MD5}$, which remains impractical to attack as far a we know.