解:(1) ①當點$P$在$y$軸的正半軸上時,過點$P'$作$P'A\perp x$軸于點$A,$
因為$P'$恰好在一次函數(shù)$y = 2x + 3$的圖象上,設(shè)$P'(m,2m + 3),$所以$P'A=-2m - 3。$
因為點$Q$的坐標為$(4,0),$所以$OQ = 4。$
因為$PQ\perp P'Q,$所以$\angle PQA+\angle AQP' = 90^{\circ},$又因為$\angle AQP'+\angle AP'Q = 90^{\circ},$所以$\angle AP'Q=\angle OQP。$
在$\triangle AP'Q$和$\triangle OQP$中,
$\begin{cases}\angle P'AQ=\angle QOP = 90^{\circ}\\\angle AP'Q=\angle OQP\\P'Q = QP\end{cases},$所以$\triangle AP'Q\cong\triangle OQP(AAS),$所以$AP' = OQ,$即$-2m - 3 = 4,$解得$m=-\frac{7}{2},$所以$P'\left(-\frac{7}{2},-4\right)。$
②當點$P$在$y$軸的負半軸上時,過點$P'$作$P'B\perp x$軸于點$B,$
因為$P'$恰好在一次函數(shù)$y = 2x + 3$的圖象上,設(shè)$P'(m,2m + 3),$所以$P'B = 2m + 3。$
同①可得$\triangle P'BQ\cong\triangle QOP,$所以$P'B = OQ,$即$2m + 3 = 4,$解得$m=\frac{1}{2},$所以$P'\left(\frac{1}{2},4\right)。$
綜上,點$P'$的坐標為$\left(-\frac{7}{2},-4\right)$或$\left(\frac{1}{2},4\right)。$