(1).设t=a^xa>1,t>0f(x)=1-2t-t^2=-(t+1)^2+2t+1>1, (t+1)^2>1f(x)<1(2).x∈[-2,1]a>1t∈[a^(-2), a]t>0f(x)对称轴t=-1,开口向下,在t=-1有最大值,在t>0递减所以在t=a处f(x)min=-7即x=1, t=a,-(t+1)^2+2=-7-(a+1)^2+2=-7a=2或-4因为a>1a=2因为t>0时f(x)递减t=a^(-2)=1/4时f(x)max=7/16