By definition,
1 parsec = Distance at which 1 AU long arc subtends an angle of 1 s.

$\begin{array}{l}\therefore 1\text{ parsec }=\left(\frac{1\mathrm{AU}}{1\text{ arc sec }}\right)\\ 1\text{ deg }=3600\text{ arc sec }\\ \therefore 1\text{ arc sec }=\frac{\mathrm{\pi }}{3600\times 180}\text{ rad }\\ \therefore 1\text{ parsec }=\frac{3600\times 180}{\mathrm{\pi }}\mathrm{AU}=206265\mathrm{AU}=2\times {10}^5\mathrm{AU}\end{array}$

Step 2: Find the angular size of the star.

(b)

Sun's diameter is ${\left(\frac{1}{2}\right)}^0$ AU.

Therefore, at 2 parsecs, the star is $\frac{1/2}{2\times 2\times {10}^5}$ degree in diameter$=0.125\times {10}^{-5}\times 60=7.5\times {10}^{-5} \mathrm{arc} \min$. With 100 magnification, it should look $7.5\times {10}^{-3}$ arc min. However, due to atmospheric fluctuations, it will still look like about 1 arcmin. It cannot be magnified using a telescope.

(c) Step 3: Find the size of Mars.

$\text{ Given that }\frac{{\mathrm{D}}_{\text{ mars }}}{{\mathrm{D}}_{\text{ eath }}}=\frac{1}{2} ...\left(\mathrm{i}\right)$

where D represents diameter.

$\mathrm{We} \mathrm{know} \mathrm{that}, \frac{{\mathrm{D}}_{\text{ earth }}}{{\mathrm{D}}_{\text{ sun }}}=\frac{1}{100}$
$\therefore \frac{{\mathrm{D}}_{\text{ mars }}}{{\mathrm{D}}_{\text{ sun }}}=\frac{1}{2}\times \frac{1}{100} \left[\mathrm{from} \mathrm{Eq}. \left(\mathrm{i}\right)\right]$
$\begin{array}{l}\text{ At }1\mathrm{AU}\text{ sun's diameter }=(\frac{1}{2}{)}^{\circ }\\ \therefore \text{ Mars diameter }=\frac{1}{2}\times \frac{1}{200}=\frac{1}{400}\\ \text{ At }\frac{1}{2}\mathrm{AU},\text{ Mars diameter }=\frac{1}{400}\times 2^{\circ }=(\frac{1}{200}{)}^{\circ }\end{array}$

With magnification, Mar's diameter $=\frac{1}{200}\times {100}^{\circ }=(\frac{1}{2}{)}^{\circ }={30}^{\mathrm{\prime }}$

This is larger than the resolution limit due to atmospheric fluctuations. Hence, it looks magnified.