= Solution
For <spin> $1/2$, \b[$\mathbf S=\hbar\boldsymbol\sigma/2$]. The <Pauli matrices> satisfy $\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k$. The antisymmetric term vanishes on contraction with $n_in_j$, and $|\mathbf n|=1$, so $(\mathbf n\cdot\boldsymbol\sigma)^2=I$. Separating the even and odd terms of the exponential series gives
$$
\boxed{e^{-i\theta\mathbf n\cdot\mathbf S/\hbar}=I\cos(\theta/2)-i(\mathbf n\cdot\boldsymbol\sigma)\sin(\theta/2).}
$$
For the single-particle <Hamiltonian>, <unitary time evolution> consequently acts as
$$
|\uparrow\rangle\mapsto\cos(\alpha t/2)|\uparrow\rangle-i\sin(\alpha t/2)|\downarrow\rangle,\qquad|\downarrow\rangle\mapsto\cos(\alpha t/2)|\downarrow\rangle-i\sin(\alpha t/2)|\uparrow\rangle.
$$
For two <spins> the joint eigenstates of total <angular momentum> are the <spin-one-half triplet state>
$$
|1,1\rangle=|\uparrow\uparrow\rangle,\quad |1,0\rangle=\frac{|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle}{\sqrt2},\quad |1,-1\rangle=|\downarrow\downarrow\rangle,
$$
with $J^2=2\hbar^2$ and $J_3=m\hbar$, and the <spin-one-half singlet state>
$$
|0,0\rangle=\frac{|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle}{\sqrt2},\qquad J^2=J_3=0.
$$
These follow by adding the two spin-$1/2$ angular momenta, or by direct action of the Pauli matrices.
The two terms of the specified <Hamiltonian> commute, so its evolution factorizes into rotations with opposite angles on the two tensor factors. The unique initial $J_3=\hbar$ state is $|\uparrow\uparrow\rangle$. With $c=\cos(\lambda t/2)$, $s=\sin(\lambda t/2)$, its evolved state is
$$
c^2|\uparrow\uparrow\rangle+s^2|\downarrow\downarrow\rangle+ics(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle).
$$
The singlet amplitude is $i\sqrt2cs$, giving the requested probability
$$
\boxed{\mathbb P(J^2=0)=2c^2s^2=\tfrac12\sin^2(\lambda t).}
$$
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