= Solution
Put $N_a=a^\dagger a$ and $N_b=b^\dagger b$. The <canonical commutation relations> give $[N_a,a^\dagger]=a^\dagger$, $[N_a,a]=-a$, and the analogous relations for $b$, with all cross-mode operators commuting. Consequently $[K_3,K_+]=K_+$ and $[K_3,K_-]=-K_-$. Also
$$
K_+K_-=N_a(N_b+1),\qquad K_-K_+=(N_a+1)N_b,
$$
so $[K_+,K_-]=N_a-N_b=2K_3$. This is the <Schwinger boson representation> of the <angular momentum> algebra, with $\hbar=1$ in the stated <commutators>.
The vacuum has $N_a=N_b=0$, and $(a^\dagger)^2|0\rangle$ has occupation numbers $(2,0)$, hence $K_3$ <eigenvalue> $1$. With $N=N_a+N_b$,
$$
K^2=N_aN_b+\frac12N+\frac14(N_a-N_b)^2
=\frac N2\left(\frac N2+1\right).
$$
Thus $\boxed{K^2(a^\dagger)^2|0\rangle=2(a^\dagger)^2|0\rangle}$. It is the highest-weight state of the spin-one <irreducible representation>. There are \b[two other states in this multiplet], $a^\dagger b^\dagger|0\rangle$ and $(b^\dagger)^2|0\rangle$, with $K_3$ <eigenvalues> $0,-1$. A normalized <basis> is $(a^\dagger)^2|0\rangle/\sqrt2$, $a^\dagger b^\dagger|0\rangle$, $(b^\dagger)^2|0\rangle/\sqrt2$. This count assumes the usual two-mode <bosonic Fock space>; an extra unmentioned vacuum multiplicity could replicate the multiplet.
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