(1)
解：因?yàn)橹本€$l:y = kx + 3$與$y$軸交于點(diǎn)$B，$所以$B(0,3)，$即$OB = 3。$
又因?yàn)?\frac{OB}{OA}=\frac{3}{4}，$所以$OA = 4，$即$A(4,0)。$
因?yàn)辄c(diǎn)$A$在直線$l$上，所以$4k+3 = 0，$$4k=-3，$解得$k=-\frac{3}{4}。$
所以直線$l$的表達(dá)式為$y = -\frac{3}{4}x+3。$
(2)
解：過點(diǎn)$P$作$PC\perp y$軸于點(diǎn)$C。$
因?yàn)?S_{\triangle BOP}=\frac{1}{2}OB\cdot PC = 6，$$OB = 3，$所以$\frac{1}{2}\times3\times PC = 6，$$PC = 4。$
所以點(diǎn)$P$的橫坐標(biāo)為$4$或$-4。$
因?yàn)辄c(diǎn)$P$為直線$l$上的一個(gè)動(dòng)點(diǎn)且不與點(diǎn)$A，$$B$重合，當(dāng)$x = 4$時(shí)，$y=-\frac{3}{4}\times4 + 3=0$（與$A$點(diǎn)重合，舍去）；
當(dāng)$x=-4$時(shí)，$y=-\frac{3}{4}\times(-4)+3 = 6。$
所以點(diǎn)$P$的坐標(biāo)為$(-4,6)$時(shí)，$\triangle BOP$的面積是$6。$
(3)
解：存在。因?yàn)?OM\perp AB$于$M，$$AB=\sqrt{OB^{2}+OA^{2}}=\sqrt{3^{2}+4^{2}} = 5，$$\angle OMP = 90^{\circ}，$$OM=\frac{OA\cdot OB}{AB}=\frac{4\times3}{5}=\frac{12}{5}。$
以$O，$$P，$$Q$為頂點(diǎn)的三角形與$\triangle OMP$全等時(shí)，$\angle OQP = 90^{\circ}。$
①當(dāng)$\triangle OMP\cong\triangle PQO$時(shí)，$PQ = OM=\frac{12}{5}，$即點(diǎn)$P$的橫坐標(biāo)為$-\frac{12}{5}$或$\frac{12}{5}。$
當(dāng)$x = -\frac{12}{5}$時(shí)，$y=-\frac{3}{4}\times(-\frac{12}{5})+3=\frac{9}{5}+3=\frac{24}{5}；$
當(dāng)$x=\frac{12}{5}$時(shí)，$y=-\frac{3}{4}\times\frac{12}{5}+3=-\frac{9}{5}+3=\frac{6}{5}。$
所以點(diǎn)$P$的坐標(biāo)為$(-\frac{12}{5},\frac{24}{5})$或$(\frac{12}{5},\frac{6}{5})。$
②當(dāng)$\triangle OMP\cong\triangle OQP$時(shí)，$OQ = OM=\frac{12}{5}，$即點(diǎn)$P，$點(diǎn)$Q$的縱坐標(biāo)為$-\frac{12}{5}$或$\frac{12}{5}。$
當(dāng)$y = -\frac{12}{5}$時(shí)，$-\frac{3}{4}x+3=-\frac{12}{5}，$$-\frac{3}{4}x=-\frac{12}{5}-3=-\frac{12}{5}-\frac{15}{5}=-\frac{27}{5}，$$x=\frac{36}{5}；$
當(dāng)$y=\frac{12}{5}$時(shí)，$-\frac{3}{4}x+3=\frac{12}{5}，$$-\frac{3}{4}x=\frac{12}{5}-3=\frac{12}{5}-\frac{15}{5}=-\frac{3}{5}，$$x=\frac{4}{5}。$
所以點(diǎn)$P$的坐標(biāo)為$(\frac{36}{5},-\frac{12}{5})$或$(\frac{4}{5},\frac{12}{5})。$
綜上所述，符合條件的點(diǎn)$P$的坐標(biāo)為$(-\frac{12}{5},\frac{24}{5})，$$(\frac{12}{5},\frac{6}{5})，$$(\frac{36}{5},-\frac{12}{5})，$$(\frac{4}{5},\frac{12}{5})。$
