Explanation

(a)
Since, $a,\ b,\ c$ are in HP, then $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in AP, let $\frac{1}{a} = p - q,\frac{1}{b} = p$ and $\frac{1}{c} = p + q,$
Where $p,q > 0\ \$and $p > q$
Now, $\frac{3a + 2b}{2a - b} + \frac{3c + 2b}{2c - b} = \frac{\frac{3}{p - q} + \frac{2}{p}\ }{\frac{2}{p - q} - \frac{1}{p}}\frac{\frac{3}{p + q} + \frac{2}{p}}{\frac{2}{p + q} - \frac{1}{p}}$
$= \frac{\frac{5p - 2q}{p(p - q)}}{\frac{p + q}{p(p - q)}} + \frac{\frac{5p + 2q}{p(p + q)}}{\frac{p - q}{p(p + q)}}$
$= \frac{5p - 2q}{p + q} + \frac{5p + 2q}{p - q}$
$= \frac{(5p - 2q)(p - q) + (5p + 2q)(p + q)}{(p^{2} + q^{2})} = 10 + \frac{14q^{2}}{p^{2} - q^{2}}$
Which is obviously greater than 10$(as\ p > q > 0)$.