= Solution
Use coordinates $(ct,x,y,z)$ and put $\beta=v/c$, $\gamma=(1-\beta^2)^{-1/2}$. For an observer moving at $+v\widehat{\mathbf x}$, the coordinate <Lorentz boost> is
$$
\Lambda=\begin{pmatrix}\gamma&-\beta\gamma&0&0\\-\beta\gamma&\gamma&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.
$$
The given <electromagnetic field tensor> transforms as $F'=\Lambda F\Lambda^{\mathsf T}$. For example,
$$
F'^{02}=\gamma F^{02}-\beta\gamma F^{12}=-\frac\gamma c(E_y-vB_z),\qquad
F'^{03}=\gamma F^{03}-\beta\gamma F^{13}=-\frac\gamma c(E_z+vB_y),
$$
while
$$
F'^{12}=-\beta\gamma F^{02}+\gamma F^{12}=-\gamma(B_z-vE_y/c^2),\qquad
F'^{13}=-\beta\gamma F^{03}+\gamma F^{13}=\gamma(B_y+vE_z/c^2).
$$
The $01$ component is unchanged because its boost <determinant> is $\gamma^2(1-\beta^2)=1$. The $23$ component is unchanged because neither index lies in the boost plane. Reading off all six components gives
$$
\boxed{\begin{aligned}
E'_x&=E_x,&B'_x&=B_x,\\
E'_y&=\gamma(E_y-vB_z),&B'_y&=\gamma(B_y+vE_z/c^2),\\
E'_z&=\gamma(E_z+vB_y),&B'_z&=\gamma(B_z-vE_y/c^2).
\end{aligned}}
$$
The signs follow the stated positive observer <velocity> and the supplied <tensor> convention. In vector form for a boost in any direction, $\mathbf E'_\parallel=\mathbf E_\parallel$, $\mathbf E'_\perp=\gamma(\mathbf E_\perp+\mathbf v\times\mathbf B)$, with analogous $\mathbf B'_\perp=\gamma(\mathbf B_\perp-\mathbf v\times\mathbf E/c^2)$.
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