Sort of. It does not improve brute force resistance in a meaningful way, but that is not the purpose either. It protects against one of the hash functions being broken.

Let us assume that we have 2 160bit hash functions, and one is broken to the degree that we can find collisions in constant time. This now means that we can break the combined hash in 2^80 rather than 2^80 + 2^80. The total brute force complexity was not improved, but the reliance on either hash function was.

Yes, but you picked two hash functions which are known to be insecure. Their complexity reduction argument appears to apply to cases where you are not using pure brute force for either hash, which would be the case if you were attacking MD5 and SHA1.

I did so quite on purpose. Both are concidered broken, but combined, only the bruteforce attack is known to work, i.e. "the whole is greater than the sum of its parts".