Question 1:
Yes, you can. Suppose that $T$ is a compact linear operator, and assume that the consequences in 5 all hold. Then, suppose that $\lambda \neq 0$ fails to be an eigenvalue (we want to show that $T - \lambda$ has a bounded inverse).
It would follow that the eigenspace $N(I - \frac 1{\lambda}T) = N(\lambda - T) = \{0\}$. By (iv), we have
$$
R(I - \frac 1{\lambda}T) = R(\lambda - T) = R(T - \lambda) = H
$$
Thus, the map $T - \lambda$ is bounded (since it is compact), is injective (since its kernel, the eigenspace, is $\{0\}$), and is surjective. By the bounded inverse theorem, $(T - \lambda)^{-1}$ exists and is bounded.
Question 2:
It seems that Evans' version contains a bit more information about compact operators. For example, nothing about Tao's version says anything about finite dimensional kernels. That being said, let's assume $X = H$ is a Hilbert space, and try to go the other direction, filling in the blanks as we go.
Suppose that $K$ is a compact operator which has $\lambda = 1$ as an eigenvalue. If $N(I - K)$ where infinite dimensional, then $K \mid_{N(I - K)}$ would be the identity operator on an infinite dimensional space, which fails to be compact. So, $N(I - K)$ must be finite dimensional.
From here, we could presumably prove that $R(I - K) = N(I - K)^{\perp}$, from which we would get (ii) for free. I'll skip that, but I expect that it's in Evans. We would also need to prove (v) without help from Tao.
Because $R(I - K)^\perp = N(I- K)^{\perp \perp} = N(I - K) \neq \{0\}$, we can conclude that $R(I - K) \neq H$.
On the other hand, suppose that $\lambda = 1$ is not an eigenvalue. Because $(T - I)^{-1}$ exists, we can conclude that $T-I$ is onto. Thus, the required conditions for (i) (ii) (iii) (iv) and (v) hold trivially. So, Tao's version certainly helps with one direction.
