= Solution
Let $L\phi=\lambda\phi$ with $\phi\ne0$ in the <self-adjoint> domain. By the boundary identity from part (i),
$$
\lambda\langle\phi,\phi\rangle=\langle\phi,L\phi\rangle=\langle L\phi,\phi\rangle=\lambda^*\langle\phi,\phi\rangle.
$$
Since its squared weighted <norm> is positive, $\boxed{\lambda\in\mathbb R}$.
For $L\phi_n=\lambda_n\phi_n$ and $L\phi_m=\lambda_m\phi_m$, the same identity gives
$$
\lambda_n\langle\phi_m,\phi_n\rangle=\lambda_m\langle\phi_m,\phi_n\rangle.
$$
Thus $\lambda_n\ne\lambda_m$ implies
$$
\boxed{\int_a^b\phi_m^*(x)\phi_n(x)w(x)\,dx=0.}
$$
This is <orthogonality of Sturm-Liouville eigenfunctions>. Degenerate <eigenspaces> can be given an <orthonormal basis>, although vectors in a degenerate <eigenspace> are not automatically orthogonal.
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