= Solution
For the <Hadamard basis>, $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt2$. The <Born rule> gives
$$
\Pr(\pm)=|\langle\pm|\phi\rangle|^2
=\frac12\left|\frac{\sqrt3}{2}\pm\frac{1+i\sqrt3}{4}\right|^2.
$$
Writing the two <computational basis> amplitudes as $a,b$, <normalization> gives $\Pr(\pm)=\frac12\pm\operatorname{Re}(a^*b)$, and $\operatorname{Re}(a^*b)=\sqrt3/8$. Hence
$$
\boxed{\Pr(+)=\frac{4+\sqrt3}{8},\qquad\Pr(-)=\frac{4-\sqrt3}{8}.}
$$
These <probabilities> sum to one. The cross term is <interference> between <probability amplitudes>, so the <probabilities> cannot be obtained by treating the input as an incoherent mixture of the <computational basis> alternatives.
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