解：?$(1)∵$?數(shù)據(jù)?$x_1，$??$x_2，$?…，?$x_6$?的平均數(shù)為?$1，$?
?$∴x_1+x_2+…+x_6=1×6=6.$?
由題意知?$\frac {1}{6}[(x_1-1)^2+(x_2-1)^2+…+(x_6-1)^2]=\frac {5}{3}，$?
?$∴\frac {1}{6}[x_1^2+x_2^2+…+x_6^2-2(x_1+x_2+…+x_6)+6]=\frac {5}{3}，$?
?$∴x_1^2+x_2^2+…+x_6^2-2(x_1+x_2+…+x_6)+6=10，$?
?$∴x_1^2+x_2^2+…+x_6^2=10-6+2(x_1+x_2+…+x_6)=10-6+12=16.$?
?$(2)$?由于平均數(shù)無變化，故?$x_7=1.$?
?$∵\(yùn)frac {1}{6}[(x_1-1)^2+(x_2-1)^2+…+(x_6-1)^2]=\frac {5}{3}，$?
?$∴(x_1-1)^2+(x_2-1)^2+…+(x_6-1)^2=10.$?
所以?$7$?個(gè)數(shù)的方差：?$\frac {1}{7}[(x_1-1)^2+(x_2-1)^2+…+(x_6-1)^2+0^2]=\frac {10}{7}.$?
