Explanation

(a)
In methane, ethene and ethyne, the hybridisations are respectively $sp^{3}$, $sp^{2}$ and $sp$. Hence, % $s$-character will be
$sp^{3} = \frac{1}{4} \times 100 = 25\%$
$sp^{2} = \frac{1}{3} \times 100 = 33\%$
$sp = \frac{1}{2} \times 100 = 50\%$