Answer
(a) The total work done is -910 J.
(b) The work done by the cyclist is 3170 J.
Work Step by Step
(a) $W_{tot} = K_2-K_1$
$W_{tot} = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$
$W_{tot} = \frac{1}{2}(80.0~kg)(1.50~m/s)^2-\frac{1}{2}(80.0~kg)(5.00~m/s)^2$
$W_{tot} = -910~J$
(b) Let $W_C$ be the work done by the cyclist.
$W_{tot} = W_g+W_C$
$W_C = W_{tot}-W_g$
$W_C = -910~J - (-(80.0~kg)(9.80~m/s^2)(5.20~m))$
$W_C = 3170~J$
The work done by the cyclist is 3170 J.