解:(1)证明:椭圆C1:x2/2+y2=1的两个焦点为F1(-1,0),F2(1,0),由题意可得λ=1/2,即椭圆C2:x2/2+y2=1/2,设P(m,n),即m2+2n2=1,AB、CD的斜率为k,k#39;,即kk#39;=n/m+1.n/m-1=n2/m2-1=n2/-2n2=-1/2,综上所述,kk#39;为定值-1/2(2)设PF1:y=k(x+1),代入椭圆方程x2+2y2=2,可得(1+2k2)x2+4k2x+2k2-2=0,设A(x1,y1),B(x2,y2),即有x1+x2=-4k2/1+2k2,x1x2=2k2-2/1+2k2,故|AB|=√1+k2.√16k4/(1+2k2)2-8k2-8/1+2k2=2√2(1+k2)/1+2k2,设PF2:y=k#39;(x-1),代入椭圆方程x2+2y2=2,可得(1+2k#39;2)x2-4k#39;2x+2k#39;2-2=0,设C(x3,y3),D(x4,y4),即有x3+x4=4k#39;2/1+2k#39;2,x3x4=2k#39;2-2/1+2k#39;2,故|CD|=√1+k#39;2.√16k#39;4/(1+2k#39;2)2-8k#39;2-8/1+2k#39;2=2√2(1+k#39;2)/1+2k#39;2,|AB|.|CD|=8.1+k2+k#39;2+(kk#39;)2/1+2(k2+k#39;2)+4(kk#39;)2=4[1+1/4(1+k2+k#39;2)],因为kk#39;=-1/2,所以k2+k#39;2≥2|kk#39;|=1,当且仅当k=k#39;=√2/2时,上式取等号,则|AB|.|CD|≤4(1+1/4×2)=9/2,综上所述,|AB|.|CD|的最大值为9/2