= Solution
Both forms have <discriminant of a binary quadratic form> $-60$. The <reduction algorithm for a positive definite binary quadratic form> gives $|b|\leq a\leq c$ and $a\leq\sqrt{60/3}<5$ for a reduced positive definite representative. Enumerating $a=1,2,3,4$ gives
$$
[1,0,15],\quad[2,2,8],\quad[3,0,5],\quad[4,2,4].
$$
The 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> $p\nmid30$, <quadratic reciprocity> gives
$$
\left(\frac{-60}{p}\right)=\left(\frac{-3}{p}\right)\left(\frac5p\right)
=\left(\frac p3\right)\left(\frac p5\right).
$$
Every <prime> $p\equiv1\pmod{15}$ 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 $2\pmod3$, so these <primes> must be represented by $f$. Similarly, <primes> $p\equiv2\pmod{15}$ have both symbols equal to $-1$, and are represented by $g$, since a <prime> represented by $f$ is $1\pmod3$. Both <residue classes> are <coprime> to 15, so <Dirichlet theorem on primes in arithmetic progressions> proves \b[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 \b[there are no <primes> represented by both forms].
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