Note: Edited from the original answer.
If you have specific values for
$n$, then the following will work for the mean and variance of the maximum for a sample of size
$n=3$:
trd = TruncatedDistribution[{a, b}, ExponentialDistribution[\[Lambda]]];
otrd = OrderDistribution[{trd, 3}, 3];
FullSimplify[Mean[otrd], Assumptions -> {b > a > 0, \[Lambda] > 0}]
$$\frac{(6 a \lambda +11) \left(-e^{3 b \lambda }\right)+18 (b \lambda +1) e^{\lambda (a+2 b)}-9 (2 b \lambda +1) e^{\lambda (2 a+b)}+e^{3 a \lambda } (6 b \lambda +2)}{6 \lambda \left(e^{a \lambda }-e^{b \lambda }\right)^3}$$
FullSimplify[Variance[otrd], Assumptions -> {b > a > 0, \[Lambda] > 0}]
$$\frac{-4 e^{3 \lambda (a+b)} (9 \lambda (a-b) (a \lambda -b \lambda +3)+155)+249 e^{2 \lambda (2 a+b)}-42 e^{\lambda (5 a+b)}-6 e^{\lambda (a+5 b)} (6 \lambda (a-b) (3 a \lambda -3 b \lambda +5)+55)+6 e^{2 \lambda (a+2 b)} (6 \lambda (a-b) (3 a \lambda -3 b \lambda +8)+115)+4 e^{6 a \lambda }+49 e^{6 b \lambda }}{36 \lambda ^2 \left(e^{a \lambda }-e^{b \lambda }\right)^6}$$
