6. 如图①,一次函数$y=2x+4$的图象与$x$轴交于点$A$,与$y$轴交于点$B$,点$C$在$x$轴的正半轴上,且$OC=2OA$.
(1)求点$A$的坐标和直线$BC$的函数表达式.
(2)如图②,设点$M$是$x$轴上的一个动点,过点$M$作$y$轴的平行线,交直线$AB$于点$P$,交直线$BC$于点$Q$.
①若$△ PQB$的面积为$\dfrac{8}{3}$,求点$M$的坐标;
②当$∠ MBC+∠ ABO=45°$时,求点$M$的坐标.

(1)对于一次函数$y=2x+4$,令$y=0$,则有$0=2x+4$,解得$x=-2$,$\therefore$点$A(-2,0)$,$\therefore OA=2$.$\because OC=2OA=2×2=4$,$\therefore C(4,0)$.对于一次函数$y=2x+4$,令$x=0$,则$y=4$,$\therefore B(0,4)$.设直线$BC$的表达式为$y=kx+b(k\ne0)$,将点$B(0,4)$,$C(4,0)$代入,可得$\begin{cases} 4=b,\\ 0=4k+b, \end{cases}$解得$\begin{cases} k=-1,\\ b=4, \end{cases}$$\therefore$直线$BC$的表达式为$y=-x+4$.
(2)①根据题意,当点$M$在$x$轴负半轴时,如图①.

可设$M(a,0)(a＜0)$,则$P(a,2a+4)$,则$Q(a,-a+4)$,$\therefore PQ=|2a+4-(-a+4)|=|3a|=-3a$.$\because △ PQB$的面积为$\dfrac{8}{3}$,$\therefore S_{△ PQB}=\dfrac{1}{2}PQ·|a|=\dfrac{1}{2}×(-3a)×(-a)=\dfrac{3}{2}a^2=\dfrac{8}{3}$,解得$a=\dfrac{4}{3}$(舍去)或$a=-\dfrac{4}{3}$.此时$M(-\dfrac{4}{3},0)$.
当点$M$在$x$轴正半轴时,如图②.可设$M(a',0)(a'＞0)$,则$P(a',2a'+4)$,则$Q(a',-a'+4)$,$\therefore PQ=|2a'+4-(-a'+4)|=|3a'|=3a'$.$\because △ PQB$的面积为$\dfrac{8}{3}$,$\therefore S_{△ PQB}=\dfrac{1}{2}PQ·|a'|=\dfrac{1}{2}×3a'× a'=\dfrac{3}{2}a'^2=\dfrac{8}{3}$,解得$a'=\dfrac{4}{3}$或$a'=-\dfrac{4}{3}$(舍去).此时$M(\dfrac{4}{3},0)$.综上所述,点$M$的坐标为$(-\dfrac{4}{3},0)$或$(\dfrac{4}{3},0)$.
②由(1)可知,$A(-2,0)$,$B(0,4)$,$C(4,0)$,$\therefore OB=OC=4$.又$\because ∠ BOC=90°$,$\therefore ∠ OBC=∠ OCB=\dfrac{1}{2}(180°-∠ BOC)=45°$.
可分两种情况讨论:当$∠ MBO=∠ ABO$时,如图③,可有$∠ MBC+∠ ABO=∠ MBC+∠ MBO=∠ OBC=45°$.在$△ ABO$和$△ MBO$中,$\begin{cases} ∠ ABO=∠ MBO,\\ BO=BO,\\ ∠ AOB=∠ MOB, \end{cases}$$\therefore △ ABO≌△ MBO(\mathrm{ASA})$,$\therefore OM=OA$,$\therefore M(2,0)$.

当$∠ M'BC=∠ MBC$时,如图④,过点$C$作$CK⊥ OM'$,交$BM'$于点$K$.可有$∠ M'BC+∠ ABO=∠ MBC+∠ MBO=∠ OBC=45°$.$\because ∠ M'BC+∠ BM'O=∠ OCB=45°$,$∠ MBC+∠ MBO=∠ OBC=45°$,$\therefore ∠ M'BC+∠ BM'O=∠ MBC+∠ MBO$,$\therefore ∠ BM'O=∠ MBO=∠ ABO$.$\because ∠ BM'O+∠ CKM'=∠ MBO+∠ OMB=90°$,$\therefore ∠ CKM'=∠ OMB$.$\because ∠ CKM'+∠ BKC=∠ OMB+∠ BMC=180°$,$\therefore ∠ BKC=∠ BMC$.在$△ BMC$和$△ BKC$中,$\begin{cases} ∠ MBC=∠ KBC,\\ ∠ BMC=∠ BKC,\\ BC=BC, \end{cases}$$\therefore △ BMC≌△ BKC(\mathrm{AAS})$,$\therefore KC=MC=4-2=2$.在$△ OBM$和$△ CM'K$中,$\begin{cases} ∠ OBM=∠ CM'K,\\ ∠ BOM=∠ M'CK=90°,\\ OM=CK=2, \end{cases}$$\therefore △ OBM≌△ CM'K(\mathrm{AAS})$,$\therefore CM'=OB=4$,$\therefore OM'=OC+CM'=4+4=8$,$\therefore M'(8,0)$.
综上所述,当$∠ MBC+∠ ABO=45°$时,点$M$的坐标为$(2,0)$或$(8,0)$.