Q:

Endianness order. How will these bytes be represented?

Given the hexadecimal bytes 0x12345678, copy the bytes to memory using big-endian order.
Address         Content
0x00400003        0x78
0x00400002        0x56
0x00400001        0x34
0x00400000        0x12
Is that right?

A:

In big-endian, the most significant byte (12) should come first, and then the rest should come in decreasing order of significance.
If the given number is in big-endian byte-order (and probably it is), your solution is right, as it will look like this:
00400000|00400001|00400002|00400003
--------+--------+--------+--------
   12   |   34   |   56   |   78

If you had to arrange the bytes in little endian, the arrangement would be reversed:
00400000|00400001|00400002|00400003
--------+--------+--------+--------
   78   |   56   |   34   |   12

Note that in this arrangement, only the order of bytes is reversed, but the order of nibbles (4-bit regions = hexadecimal digits) remains the same.
You can read more in this Wikipedia page about endianness.

