= Solution
An endomorphism $\alpha$ is a projection onto its image precisely when it is the identity on $\operatorname{im}\alpha$ and kills a complementary subspace. If $\alpha^2=\alpha$, then
$$
\alpha(\alpha v)=\alpha v,
$$
so $\alpha$ is the identity on its image. Moreover,
$$
v=\alpha v+(v-\alpha v),
\qquad
\alpha(v-\alpha v)=0,
$$
and $\operatorname{im}\alpha\cap\ker\alpha=\{0\}$. Hence
$$
V=\operatorname{im}\alpha\oplus\ker\alpha,
$$
and $\alpha$ is the projection onto its image along its kernel. Conversely, every projection is the identity after one application, so applying it twice has the same effect: $\alpha^2=\alpha$.
Statement (i) is \b[false]. On a two-dimensional vector space, take the nonzero nilpotent endomorphism
$$
\alpha=
\begin{pmatrix}
0&1\\
0&0
\end{pmatrix}.
$$
Then $\alpha^2=\alpha^3=0$, but $\alpha^2\ne\alpha$, so $\alpha$ is not <idempotent>.
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