As Tomas points out, $P(a) = P(b) = P(c) = \frac{1}{3}$. Thus,
$E(X_1) = aP(a) + bP(b) + cP(c)$
$=\frac{a+b+c}{3}$
You can apply the same method to work out $E(X_2)$.
As Vincent points out, the linearity of Expected Value over these uniform distributions gives
$E(X_1+X_2) = E(X_1) + E(X_2)$
[Alternatively, you could go through all the possibilities of $X_1+X_2$, i.e. the set {$a+d, a+e, b+d,b+e,c+d,c+e$}. There are six elements in this set and the probability of getting each is $\frac{1}{6}$, because the distribution is uniform. And you could compute $E(X_1+X_2)$ in a similar fashion as above.]