If $A^k = I$, then we have $A^k - I = 0$. Factoring, we have
$$
(A - I)(I + A + A^2 + \cdots + A^{k-1}) = 0
$$
Since $1$ is not an eigenvalue, $A-I$ is invertible so that
$$
I + A + A^2 + \cdots + A^{k-1} = 0
$$
In fact, $A$ satisfies the hypothesis of your question if and only if the above equality holds. Let $p(x)$ denote the polynomial
$$
p(x) = 1 + x + x^2 + \cdots + x^{k-1}
$$
1) necessarily holds. Note that $p(x)$ factors as
$$
p(x) = \prod_{j=1}^{k-1} (x - e^{2 \pi ij/k})
$$
since the minimal polynomial of $A$ divides $p$, we may conclude that $A$ is diagonalizable.
2) Never holds. In particular, since $p(A) = 0$, we have
$$
A + A^2 + \cdots + A^{k-1} = -I
$$
3) holds by the above equality
4) is true. Note that $A$ is necessarily invertible, and that
$$
A + A^2 + \cdots + A^{k-1} = -I \implies\\
A^{-k}(A + A^2 + \cdots + A^{k-1}) = -A^{-k} \implies\\
A^{-1} + A^{-2} + \cdots + A^{-k+1} = -A^{-k} = -[A^{k}]^{-1} = -I^{-1} = -I
$$\