= Solution
For the <multiplication operator>, $\|Tf\|_\infty\leq\|g\|_\infty\|f\|_\infty=\|f\|_\infty$. Taking $f=1$ gives equality, so $\|T\|=1$.
An <eigenfunction> obeys $(g(t)-\lambda)f(t)=0$ at every $t$. If $\lambda\in[0,1)$, its level set is the single point $t=\lambda$, and a <continuous function> supported at a single point is zero. If $\lambda\notin[0,1]$, that level set is empty. For $\lambda=1$, every <continuous function> vanishing on $[0,1]$ is an <eigenfunction>, and nonzero examples exist, such as $f(t)=\max\{t-1,0\}$. Thus $\sigma_p(T)=\{1\}$.
If $\lambda\notin[0,1]$, multiplication by $(g-\lambda)^{-1}$ is a bounded inverse of $T-\lambda I$. If $\lambda\in[0,1]$, choose $t_0$ with $g(t_0)=\lambda$. Every function in the range of $T-\lambda I$ vanishes at $t_0$, so the range omits the constant function one. Hence it is not onto. Therefore
$$
\boxed{\|T\|=1,\qquad\sigma_p(T)=\{1\},\qquad\sigma(T)=[0,1].}
$$
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