【解答】解:(1)如圖1,點(diǎn)Q在線段AB上,設(shè)Q(a,b).過點(diǎn)Q作QC⊥OB于點(diǎn)C,過點(diǎn)Q作QD⊥OA于點(diǎn)D.∵點(diǎn)A(8,0),點(diǎn)B(0,6).∴OB=6,OA=8.∴在Rt△AOB中,根據(jù)勾股定理求得AB=10.∵CQ∥OA,∴∠1=∠2,∴cos∠1=cos∠2,即OAAB=CQBQ,∴810=a2t-6,解得,a=8t-245.又∵sin∠2=OBAB=b10-(2t-6),即610=b16-2t,解得b=48-6t5,∴Q點(diǎn)坐標(biāo)為(8t-245,48-6t5);(2)如圖1,當(dāng)t>3時(shí),點(diǎn)Q在線段AB上.由(1)知,OD=a=8t-245∴PD=OP-OD=t-a=24-3t5,又由(1)知,QD=b=48-6t5,∴tan∠QPO=QDPD=48-6t524-3t5=2,即tan∠QPO=2;(3)當(dāng)點(diǎn)Q在OB邊上運(yùn)動(dòng)時(shí),△OQP總是直角三角形,此時(shí)0<t≤3;當(dāng)點(diǎn)Q在邊BA上運(yùn)動(dòng)時(shí),如圖1,只有∠OQP=90°,過Q點(diǎn)作QH⊥OA,垂足為H,則tan∠QPO=tan∠OQH=OHQH=2,∴8t-245:48-6t5=2,解得t=6.∴當(dāng)0<t≤3或t=6時(shí),△OQP是直角三角形;(4)當(dāng)OQ=PQ時(shí),易求t=4811;當(dāng)OQ=OP時(shí),如圖3,過O點(diǎn)作OM⊥PQ,垂足為M;過Q點(diǎn)作QH⊥OP,垂足為H.設(shè)HP=x,則QH=2x,QP=5x,QM=PM=52x,OM=5x,OP=52x,OH=32x,∴OH:OP=3:5,8t-245:t=3:5解得t=4.8.當(dāng)t=4811或4.8時(shí),△OPQ是以O(shè)Q為腰的等腰三角形.