(Edited: I seemed to have forgotten an $\epsilon$ in my earlier answer and would now say)
$\hat{x}_1 = \frac{ 4 }{5}$, $\hat{x}_2 = \frac{ 4 }{5}$, and $\hat{x}_3 = 0 $, with $\hat{\epsilon}_1 = \frac{ 1 }{5}$, $\hat{\epsilon}_2 = \frac{ 1 }{5}$, and $\hat{\epsilon}_3 = 0 $.
Generalising further with some justification added as a response to a comment:
This stat.SE question and answer gives an explanation of how to derive this. We will take all means to be zero.
The slightly harder part is estimating $Y_1$,$Y_2$ and $Y_3$. They each have variance $\sigma_y^2=\sigma_x^2+\sigma_{\epsilon}^2 $. So $S_1$ and $S_2$ each have variance $2\sigma_y^2$ and have covariance $\sigma_y^2$. Let's define
$T_1 = 2Y_1$ $T_2=Y_2+Y_3$ $T_3 = S_1+S_2 = 2Y_1 +Y_2+Y_3=T_1+T_2 $ $T_4=S_1-S_2=Y_2-Y_3$
so $T_1$ has variance $4\sigma_y^2$ and $T_2$ has variance $2\sigma_y^2$ and they are uncorrelated. This means $T_3$ has variance $6\sigma_y^2$ and $T_4$ has variance $4\sigma_y^2$ and they too are uncorrelated.
The observation of $s_1$ and $s_2$ tells us now that $t_3 = s_1 + s_2$. From this we can get maximum likelihood estimates of $t_1$ and $t_2$ in proportion to their variances so
$\hat{t}_1 = \frac{2(s_1 + s_2)}{3}$ $\hat{t}_2 = \frac{s_1 + s_2}{3}$
but since $2Y_1 = T_1$ this gives
$\hat{y}_1 = \frac{s_1 + s_2}{3}$
and we also have from the observation $t_4 = s_1 - s_2$, and from the definitions we have $2Y_2=T_2+T_4$ and $2Y_3=T_2-T_4$, giving us
$\hat{y}_2 = \frac{2s_1-s_2}{3} $ $\hat{y}_3 = \frac{2s_2-s_1}{3} .$
Getting from $\hat{y}_i$ to $\hat{x}_i$ is easier, since from the stat.SE question, we simply do this in proportion to variance, so $\hat{x}_i = \hat{y}_i \frac{\sigma_x^2}{\sigma_x^2+\sigma_{\epsilon}^2}$ giving
$\hat{x}_1 = \frac{(s_1 + s_2)\sigma_x^2}{3(\sigma_x^2+\sigma_{\epsilon}^2)}$ $\hat{x}_2 = \frac{(2s_1-s_2)\sigma_x^2}{3(\sigma_x^2+\sigma_{\epsilon}^2)} $ $\hat{x}_3 = \frac{(2s_2-s_1)\sigma_x^2}{3(\sigma_x^2+\sigma_{\epsilon}^2)} $
and
$\hat{\epsilon}_i = \hat{x}_i \frac{\sigma_{\epsilon}^2}{\sigma_x^2}.$