a. The digit after each 2 must be a non-2, and after each non-2 must be a 2, so 2's and non-2's alternate.

b. The 1's and 3's, the count of which adds to 5, can be considered like a binary number of 5 bits. Taken by themselves there are 2^5 possibilities. But there's also the choice of whether to start the number with the first non-2, or to start with the first 2, so altogether there are 2^6 cute numbers.

c. The sum of all cute numbers needs to be a multiple of 1111111111, as each digit position's value in the sum, without regard to "carries", is the same. That number, 1111111111, is divisible by 11, the only prime factor of 1408 other than 2 (1408 = 2^7 * 11). So now we need only show that the sum of any individual position's values is a multiple of 2^7.

Half of the 2^6 cute numbers have a 2 in any given position. The other half are evenly divided between 1 and 3; the total of the 1's and 3's is the same as if they were all 2's. So there are the equivalent of 2^6 2's being added together, for a total of 2^7, precisely what we said we needed.  QED