= Solution
Fix coordinates $x^\mu=(ct,x,y,z)$ and use the following convention for the contravariant <electromagnetic field tensor>:
$$
F^{\mu\nu}=\begin{pmatrix}
0&-E_x/c&-E_y/c&-E_z/c\\
E_x/c&0&-B_z&B_y\\
E_y/c&B_z&0&-B_x\\
E_z/c&-B_y&B_x&0
\end{pmatrix}.
$$
For $O'$ moving with <velocity> $+v\widehat{\mathbf x}$ relative to $O$, the <Lorentz transformation> has
$$
\Lambda=\begin{pmatrix}
\gamma&-\beta\gamma&0&0\\
-\beta\gamma&\gamma&0&0\\
0&0&1&0\\
0&0&0&1
\end{pmatrix},\qquad
\beta=\frac vc,\quad\gamma=(1-\beta^2)^{-1/2}.
$$
The rank-two <tensor> transformation rule is
$$
F'^{\mu\nu}=\Lambda^\mu{}_{\alpha}\Lambda^\nu{}_{\beta}F^{\alpha\beta},\qquad
F'=\Lambda F\Lambda^{\mathsf T}.
$$
For example,
$$
F'^{02}=\gamma F^{02}-\beta\gamma F^{12}
=-\frac\gamma c(E_y-vB_z),
$$
$$
F'^{12}=-\beta\gamma F^{02}+\gamma F^{12}
=-\gamma\left(B_z-\frac v{c^2}E_y\right).
$$
Similarly, $F'^{03}=-\gamma(E_z+vB_y)/c$ and $F'^{13}=\gamma(B_y+vE_z/c^2)$. The $01$ component is unchanged because $\gamma^2(1-\beta^2)=1$, and the $23$ component is unchanged since neither index lies in the boost plane. Reading off the six field components gives
$$
\boxed{\begin{aligned}
E'_x&=E_x,& B'_x&=B_x,\\
E'_y&=\gamma(E_y-vB_z),&B'_y&=\gamma(B_y+vE_z/c^2),\\
E'_z&=\gamma(E_z+vB_y),&B'_z&=\gamma(B_z-vE_y/c^2).
\end{aligned}}
$$
An alternative overall sign convention for $F^{\mu\nu}$ gives the same physical <Lorentz transformation of electromagnetic fields> when its identifications with $\mathbf E$ and $\mathbf B$ are used consistently.
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