$$f: Z \longrightarrow Z_7, f(k)=[k]$$
$$f: Z \longrightarrow Z_8, f(k)=[k]$$
I think this can't be one-to-one at all since it repeats after 6, and since each $f(k)$ value has a value $k$, then it is onto. However, it's the same reason for these two functions, so I'm not sure if there is something different about them aside from the modulo.
edit: To be a little more clearer, he provides these functions as well:
$$f: Z \longrightarrow Z_7, f(k)=[2k]$$
$$f: Z \longrightarrow Z_8, f(k)=[2k]$$
These two are different since it is multiples of 2, so some classes cannot be reached, such as [1], so it is neither one-to-one nor onto.
Almost. Three of your four answers are right. But you should examine the map from $\mathbb{Z}$ to $\mathbb{Z}_7$ that sends $k$ to $ 2k \pmod{7}$ a little more carefully. It's small enough to compute by brute force.