= Solution
Because $\overline\Omega$ is compact, $\phi$ attains a maximum. If a maximizer lay in $\Omega$, its <Hessian matrix> would be negative semidefinite, so its trace would satisfy $\Delta\phi\leq0$, contradicting $\Delta\phi>0$. Hence every strictly <subharmonic function> under the stated hypotheses attains a maximum on $\partial\Omega$.
Statement (i) is true. For $\epsilon>0$, the function
$$
\phi_\epsilon(x)=\phi(x)+\epsilon|x|^2
$$
has $\Delta\phi_\epsilon=4\epsilon>0$, so
$$
\max_{\overline\Omega}\phi_\epsilon
=\max_{\partial\Omega}\phi_\epsilon.
$$
Letting $\epsilon\downarrow0$ gives
$$
\boxed{\max_{\overline\Omega}\phi
=\max_{\partial\Omega}\phi}.
$$
This is the <maximum principle for harmonic functions>.
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