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解：(1) 由等比数列性质 a3*q^5 = a8
所以 q^5 = a8/a3 = 32
故q = 2; a1 = a3/q^2 = 1/4
故通项公式为 an = 2^n/4 = 2^(n-2)
(2) 由等差数列性质 b6 = b4+2d;带入等式有
b4+b4+2d = 10; b4(b4+2d) = 16;
解得：b4 = 2 d = 3（舍去d=-3）
故通项公式为 bn = b4 +(n-4)d= 2+3n -12 = 3n-10
(3)b4 = 2;am = 2^(m-2) = 2; 解得m = 3;
b6 = 8；ak = 2^(k-2) = 8 解得k = 5
所以 q^5 = a8/a3 = 32
故q = 2; a1 = a3/q^2 = 1/4
故通项公式为 an = 2^n/4 = 2^(n-2)
(2) 由等差数列性质 b6 = b4+2d;带入等式有
b4+b4+2d = 10; b4(b4+2d) = 16;
解得：b4 = 2 d = 3（舍去d=-3）
故通项公式为 bn = b4 +(n-4)d= 2+3n -12 = 3n-10
(3)b4 = 2;am = 2^(m-2) = 2; 解得m = 3;
b6 = 8；ak = 2^(k-2) = 8 解得k = 5
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