 Solution

Both forms have discriminant of a binary quadratic form $-60$. The reduction algorithm for a positive definite binary quadratic form gives $|b|\le a\le c$ and $a\le \sqrt{60/3}<5$ for a reduced positive definite representative. Enumerating $a=1,2,3,4$ givesThe second and fourth forms are imprimitive and represent only even integers. Every odd prime represented by a form of discriminant of a binary quadratic form $-60$ must therefore be represented by $f$ or $g$.

For a prime $p\nmid 30$, quadratic reciprocity givesEvery prime $p\equiv 1(\mathrm{mod}15)$ has this symbol equal to one and hence is represented by one of the two primitive forms, by part (b). But an odd prime represented by $g$ and different from three is $2(\mathrm{mod}3)$, so these primes must be represented by $f$. Similarly, primes $p\equiv 2(\mathrm{mod}15)$ have both symbols equal to $-1$, and are represented by $g$, since a prime represented by $f$ is $1(\mathrm{mod}3)$. Both residue classes are coprime to 15, so Dirichlet theorem on primes in arithmetic progressions proves each form represents infinitely many primes.

The same modulo-three distinction excludes every prime other than three from being represented by both. The form $f$ does not represent three; $g$ represents three and five but neither is represented by $f$, and two is represented by neither. Therefore there are no primes represented by both forms.