Answer
a) $58.8\;\rm N$
b) $67.7^\circ $
c) $79\;\rm N$
Work Step by Step
First of all, we need to draw the force diagram exerted on the foot and the hanging mass, as you see in the figures below.
a) From the force diagram of the mass, the net force exerted on it is zero since the whole system is in equilibrium.
$$\sum F_{y,mass}=T-mg=0$$
Thus, the tension in the rope is given by
$$T=mg=6\cdot 9.8=\color{red}{\bf 58.8}\;\rm N$$
b) We know that the patient foot is in equilibrium, so the net force exerted on it in both directions is zero.
We need to analyze the tension forces into their components, as you see below.
The traction pull force needed is given by
$$\sum F_{x,foot}=T\cos\theta+T\cos15^\circ -F_{pull}=0$$
Thus,
$$F_{pull}=T\cos\theta+T\cos15^\circ $$
Plugging the known;
$$F_{pull}=58.8\cos\theta+58.8\cos15^\circ \tag 1$$
$$\sum F_{y,foot}=T\sin\theta-T\sin15^\circ-F_G=0$$
$$T\sin\theta-T\sin15^\circ-F_G=0$$
$$58.8\sin\theta-58.8\sin15^\circ-m_{foot}g=0$$
$$58.8\sin\theta=58.8\sin15^\circ+(4\cdot 9.8) $$
Solving for $\theta$;
$$\theta=\sin^{-1}\left[\dfrac{58.8\sin15^\circ+(4\cdot 9.8) }{58.8}\right]=\color{red}{\bf 67.7^\circ }$$
c) Plugging into (1);
$$F_{pull}=58.8\cos67.7^\circ+58.8\cos15^\circ =\color{red}{\bf 79}\;\rm N $$
