= Solution
Put $\Omega=\lambda B$, including its sign. The initial vector is the $+\hbar/2$ <eigenstate> of $S_z$; it is not an <energy eigenstate> of $H=\Omega S_x$. In terms of the <spin eigenstates> found above, $|+z\rangle=(|+x\rangle+|-x\rangle)/\sqrt2$. Each <energy eigenstate> acquires its own <unitary time evolution> phase:
$$
|\psi(t)\rangle=\frac{e^{-i\Omega t/2}|+x\rangle+e^{i\Omega t/2}|-x\rangle}{\sqrt2}=\boxed{\begin{pmatrix}\cos(\Omega t/2)\\-i\sin(\Omega t/2)\end{pmatrix}.}
$$
Equivalently, $e^{-iHt/\hbar}=I\cos(\Omega t/2)-i\sigma_x\sin(\Omega t/2)$ follows from $\sigma_x^2=I$. The <Born rule> gives
$$
\boxed{P(S_z=+\hbar/2;t)=|\langle+z|\psi(t)\rangle|^2=\cos^2(\Omega t/2).}
$$
The <expectation value> of the <spin angular momentum> is $(\hbar/2)(0,-\sin\Omega t,\cos\Omega t)$, a rotation around the $x$ axis. Its direction follows the specified sign of the <Hamiltonian (quantum mechanics)>; changing the magnetic coupling sign reverses the precession.
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