（Ⅰ）2-

1

2

+

(-4)0

2

+

1

2 -1

-

(1- 5 )0

=

2

2

+

2

2

+

2

+1-1

=2

2

．

（Ⅱ）2×(lg

2

)2+

1

2

lg2×lg5+

(lg 2 )2-lg2+1

=2×（

1

2

lg2）²+

1

2

lg2•lg5+

(lg 2 -1)2

=

1

2

lg²2+

1

2

lg2•lg5-（lg

2

-1）

=

1

2

lg22+

1

2

lg2•lg5-

1

2

g2+1

=

1

2

lg2(lg2+lg5-1)+1

=

1

2

lg2(lg10-1)+1

=1．