(a) Let $\theta$ angle subtended at distance $r$ due to an arc of length $l$ is
$$
\theta=\frac{l}{r}
$$
Given, $l=R_E ; r=60 R_E$
$$
\begin{aligned}
\theta &=\frac{\operatorname{arc}(l)}{\operatorname{radius}(r)}=\frac{R_E}{60 R_E} \\
&=\frac{1}{60} \mathrm{rad}=\frac{1}{60} \times \frac{180}{\pi} \text { degree } \\
\theta &=\frac{3}{\pi}=1^{\circ}
\end{aligned}
$$
Then, the angle subtended from the moon to diameter of the earth
$$
=2 \theta=\frac{2 \times 3}{\pi}=\frac{6}{\pi} \simeq 2^{\circ}
$$
(b) Given that, as moon is seen from earth angle $\left(\frac{1}{2}\right)^{\circ}$ diameter and if earth is seen as $2^{\circ}$ diameter.
So, $\frac{\text { Diameter of earth }}{\text { Diameter of moon }}=\frac{(2 / \pi) \mathrm{rad}}{\left(\frac{1}{2 \pi}\right) \mathrm{rad}}=4$
i.e. diameter of moon/diameter of earth $=\frac{1}{4}$.
(c) From parallel measurement the sun is at a distance of about 400 times the earth-moon distance, so, $\frac{r_{\text {sun }}}{r_{\text {moon }}}=400$ (given)
(where, $r$ stands for distance and $D$ for diameter) sun and moon both appear to be of the same angular diameter as seen from the earth.
$$
\begin{aligned}
&\therefore \frac{D_{\text {sun }}}{r_{\text {sun }}}=\frac{D_{\text {moon }}}{r_{\text {moon }}} \text { or } \frac{D_{\text {sun }}}{D_{\text {moon }}}=\frac{r_{\text {sun }}}{r_{\text {moon }}} \\
&\therefore \frac{D_{\text {sun }}}{D_{\text {moon }}}=400 \quad\left(\because \frac{r_{\text {sun }}}{r_{\text {moon }}}=400 \text { given }\right) \\
&\text { But, } \frac{D_{\text {earth }}}{D_{\text {moon }}}=4 \quad(\because \text { from (b)) } \\
&\therefore \frac{D_{\text {sun }}}{D_{\text {earth }}}=100 \\
&D_{\text {sun }}=100 D_{\text {earth }}
\end{aligned}
$$