（1）2(3+1)(3^2+1)(3^4+1)...(3^32+1)+2
=(3-1)(3+1)(3^2+1)(3^4+1)...(3^32+1)+2
=(3^2-1)(3^2+1)(3^4+1)...(3^32+1)+2
=(3^4-1)(3^4+1)...(3^32+1)+2
...
=(3^64-1)+2
=3^64+1
=(81)^16+1
因(81)^16的个位数字是1
所以(81)^16+1的个位数字是2
（2）由x=3时ax^5+bx^3+cx-5的值为7得
3^5a+3^3b+3c=12
所以,当x=-3时
ax^5+bx^3+cx-5
=-(3^5a+3^3b+3c)-5
=-12-5
=-17