= Solution
For $K=\mathbb R$, the answer is yes. A root of $\Phi_4(X)=X^2+1$ is $i$ or $-i$, so
$$
\mathbb R(\zeta_4)=\mathbb C=\overline{\mathbb R}.
$$
For $K=\mathbb Q$, the answer is no. Every $\mathbb Q(\zeta_n)/\mathbb Q$ is finite abelian Galois. If $\sqrt[3]{2}$ belonged to $\mathbb Q^{\mathrm{cyc}}$, then it would belong to some $\mathbb Q(\zeta_n)$. The <subextensions of an abelian Galois extension> result would make
$$
\mathbb Q(\sqrt[3]{2})/\mathbb Q
$$
Galois. This is false: $X^3-2$ has two nonreal roots absent from the real field $\mathbb Q(\sqrt[3]{2})$. Hence $\mathbb Q^{\mathrm{cyc}}$ is a proper subfield of $\overline{\mathbb Q}$.
For $K=\mathbb F_p$, the answer is yes. Given $m\geq1$, take
$$
n=p^m-1,
$$
which is coprime to $p$. Every root of $\Phi_n$ in characteristic $p$ has exact multiplicative order $n$. The degree
$$
[\mathbb F_p(\zeta_n):\mathbb F_p]
$$
is the least positive $d$ such that $n\mid p^d-1$. The value $d=m$ works, and no $d<m$ can work because
$$
0<p^d-1<p^m-1=n.
$$
Therefore
$$
\mathbb F_p(\zeta_{p^m-1})=\mathbb F_{p^m}.
$$
Every element algebraic over $\mathbb F_p$ lies in some finite field $\mathbb F_{p^m}$, so
$$
\mathbb F_p^{\mathrm{cyc}}=\overline{\mathbb F}_p.
$$
These are respectively the <maximal cyclotomic extension of the real numbers>, the <maximal cyclotomic extension of the rational numbers>, and the <maximal cyclotomic extension of a finite field>.
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