= Solution
From $Te_0=e_0\ne0$, one is an <eigenvalue>. Zero is also an <eigenvalue>: the nonzero <continuous function> $q(x)=x(1-x)$ has both endpoint values zero, so $Tq=0$.
For $\lambda\notin\{0,1\}$, the <linear projection> identity $T^2=T$ gives an explicit bounded inverse:
$$
(\lambda I-T)^{-1}=\frac1\lambda(I-T)+\frac1{\lambda-1}T.
$$
Multiplying on either side by $\lambda I-T$ yields $(I-T)+T=I$. There are no other points in the <spectrum>, and thus the <spectrum of a nontrivial bounded projection> here is
$$
\boxed{\sigma(T)=\{0,1\},\qquad r(T)=1.}
$$
Let $M=\|T\|$, as calculated in part (ii). Its endpoint values ensure $1\leq M<\infty$. Since $T^j=T$ for all $j\geq1$,
$$
\|T^j\|^{1/j}=M^{1/j}\longrightarrow1,
\qquad\inf_{j\geq1}\|T^j\|^{1/j}=1.
$$
This verifies the <spectral radius formula> explicitly:
$$
\boxed{r(T)=\inf_{j\geq1}\|T^j\|^{1/j}=\lim_{j\to\infty}\|T^j\|^{1/j}=1.}
$$
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