= Solution
Irreducibility of the <cyclotomic polynomial> gives $[K:\mathbb Q]=\varphi(15)=8$. The extension is <Galois>, since $K$ is the <splitting field> of $T^{15}-1$ in characteristic zero. The injection in part (a) is consequently an isomorphism, and the <Chinese remainder theorem> gives
$$
\operatorname{Gal}(K/\mathbb Q)\cong(\mathbb Z/15\mathbb Z)^\times\cong C_2\times C_4.
$$
The <Fundamental theorem of Galois theory> identifies proper intermediate fields strictly above $\mathbb Q$ with proper nontrivial <subgroups>. Such subgroups have order two or four. The two-torsion subgroup is $C_2\times C_2$, with three nonidentity elements and hence three order-two subgroups. There are also three order-four subgroups: they are the kernels of the three nonzero <group homomorphisms> $C_2\times C_4\to C_2$. Distinct such maps have distinct kernels because the target has only one nonidentity element. Therefore
$$
\boxed{\text{three quadratic subfields and three quartic subfields, six in total}.}
$$
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