= Solution
{c}
The <angular momentum eigenstates> have $J^2$ <eigenvalue> $\hbar^2j(j+1)$, where the <angular momentum quantum number> is $j=0,\tfrac12,1,\tfrac32,\ldots$. For each $j$, the <magnetic quantum number> runs in unit steps from $-j$ to $j$. Here that <magnetic quantum number> is $+1/2$, so only half-integer $j$ are compatible:
$$
\boxed{\lambda=j(j+1),\qquad j=\frac12,\frac32,\frac52,\ldots,\qquad
\lambda\in\left\{\frac34,\frac{15}4,\frac{35}4,\ldots\right\}.}
$$
Integer $j$ cannot contain a $J_3$ <eigenvalue> $\hbar/2$. Applying the <angular momentum lowering operator> and using $[J_3,J_-]=-\hbar J_-$ gives
$$
J_3\varphi=J_3J_-\psi=(J_-J_3-\hbar J_-)\psi
=\boxed{-\frac\hbar2\varphi}.
$$
The lowered vector is nonzero, since the <adjoint operator> identity $J_-^*=J_+$ gives
$$
\|\varphi\|^2=\langle\psi,J_+J_-\psi\rangle
=\hbar^2\left(\lambda-\frac14+\frac12\right)
=\hbar^2\left(\lambda+\frac14\right)>0.
$$
Finally, $J_3$ is a <self-adjoint operator>, so its <eigenstates> for distinct <eigenvalues> are <orthogonal>. Explicitly,
$$
-\frac\hbar2\langle\varphi\mid\psi\rangle
=\langle J_3\varphi\mid\psi\rangle
=\langle\varphi\mid J_3\psi\rangle
=\frac\hbar2\langle\varphi\mid\psi\rangle,
$$
and hence $\boxed{\langle\varphi\mid\psi\rangle=0}$.
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