Answer
See the detailed answer below.
Work Step by Step
First, we need to draw the force diagram of the object.
As we see in the last figure below, we analyzed the forces exerted on the crate into its $x$ and $y$ components.
a) The work done by tension is given by
$$W_{T}=T\cos18^\circ \Delta x$$
where $\Delta x$ is the distance traveled up the incline by the carte.
Plugging the known;
$$W_{T}=120\cos18^\circ \times 5 =\color{red}{\bf 571}\;\rm J$$
The work done by the gravity is given by
$$W_{F_G}=-mg\sin30^\circ \Delta x$$
The negative sign is due to the direction of the gravitational force component.
Plugging the known;
$$W_{F_G}=-(8)(9.8)\sin30^\circ \times 5=\color{red}{\bf -196}\;\rm J$$
The work done by the normal force is given by
$$W_n=F_n\cos90^\circ \Delta x=\color{red}{\bf 0}\;\rm J$$
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b)
The amount of thermal energy is given by the work done by the kinetic friction force.
$$\Delta E_{th}=W_{f_k}=f_k \Delta x$$
We know that $f_k=\mu_kF_n$
$$\Delta E_{th} = \mu_kF_n \Delta x\tag 1$$
So we have to found the normal force which is given by applying Newton's second law in $y$-direction.
$$\sum F_y=F_n+T\sin18^\circ -mg\cos30^\circ=ma_y=m(0)=0$$
Thus,.
$$F_n=mg\cos30^\circ-T\sin18^\circ $$
Plugging into (1);
$$\Delta E_{th} = \mu_k(mg\cos30^\circ-T\sin18^\circ)\Delta x $$
Plugging the known;
$$\Delta E_{th} = (0.25)(8 \times9.8\cos30^\circ-120\sin18^\circ)(5) $$
$$\Delta E_{th} =\color{red}{\bf 38.52}\;\rm J$$
