Just perform the computation using: $R(t) = p \\{\omega | T(\omega) > t \\} = \int_t^\infty f(x) dx$.
It may help to draw $f$ and think about what $R(t)$ means in terms of the graph of $f$.
It should be clear that no devices fail for $tb$. This tells you what $R(t)$ is for these ranges.
Remember that $f$ is a pdf.
Consider three cases:
(1) $t
> We have $R(t) = \int_t^\infty f(x) dx = \int_a^b f(x) dx = 1$.
(2) $t \in [a,b]$:
> $R(t) = \int_t^b f(x) dx = {b -t \over b-a}$.
(3) $t >b$:
> $R(t) = \int_t^\infty f(x) dx = 0$.