解：?$(1)$?因?yàn)閿?shù)據(jù)?$x_{1}，x_{2}，....，x_{6}$?的平均數(shù)為?$1，$?
所以?$x_{1} + x_{2}+.... + x_{6}=1×6 = 6。$?
?$ $?又因?yàn)檫@組數(shù)據(jù)的方差為?$\frac {5}{3}，$?
根據(jù)方差公式?$s^2=\frac {1}{n}[(x_{1}-\overline {x})^2+(x_{2}-\overline {x})^2+....+(x_{n}-\overline {x})^2]($?這里?$n = 6，$??$\overline {x}=1)，$?
可得?$(x_{1} - 1)^2+(x_{2} - 1)^2+....+(x_{6} - 1)^2=\frac {5}{3}×6 = 10。$?
?$ $?將?$(x_{1} - 1)^2+(x_{2} - 1)^2+...+(x_{6} - 1)^2$?展開：
?$ \begin {aligned}&(x_{1} - 1)^2+(x_{2} - 1)^2+....+(x_{6} - 1)^2\\=&x_{1}^2-2x_{1} + 1+x_{2}^2-2x_{2} + 1+....+x_{6}^2-2x_{6} + 1\\=&x_{1}^2+x_{2}^2+....+x_{6}^2-2(x_{1} + x_{2}+....+ x_{6})+6\end {aligned}$?
?$ $?所以?$x_{1}^2+x_{2}^2+....+x_{6}^2-2×6 + 6=10，$?即?$x_{1}^2+x_{2}^2+....+x_{6}^2=16。$?
?$(2)$?因?yàn)閿?shù)據(jù)?$x_{1}，x_{2}，....，x_{7}$?的平均數(shù)為?$1，$?
所以?$x_{1} + x_{2}+....+ x_{7}=1×7 = 7。$?
?$ $?又因?yàn)?$x_{1} + x_{2}+....+ x_{6}=6，$?
所以?$6 + x_{7}=7，$?
解得?$x_{7}=1。$?
?$ $?這?$7$?個(gè)數(shù)據(jù)的方差為：
?$ \begin {aligned}s^2&=\frac {1}{7}[(x_{1} - 1)^2+(x_{2} - 1)^2+·s+(x_{6} - 1)^2+(x_{7} - 1)^2]\\&=\frac {1}{7}[10+(1 - 1)^2]\\&=\frac {10}{7}\end {aligned}$?
