(A) Average atomic mass of magnesium,$m = 24.312 \; u$.
Mass of $_{12}^{24} Mg$ isotope,$m_1 = 23.98504 \; u$.
Mass of $_{12}^{25} Mg$ isotope,$m_2 = 24.98584 \; u$.
Mass of $_{12}^{26} Mg$ isotope,$m_3 = 25.98259 \; u$.
Abundance of $_{12}^{24} Mg$,$\eta_1 = 78.99 \%$.
Let the abundance of $_{12}^{25} Mg$ be $\eta_2 = x \%$.
Then,the abundance of $_{12}^{26} Mg$ is $\eta_3 = (100 - 78.99 - x) \% = (21.01 - x) \%$.
The average atomic mass is given by $m = \frac{m_1 \eta_1 + m_2 \eta_2 + m_3 \eta_3}{100}$.
Substituting the values: $24.312 = \frac{23.98504 \times 78.99 + 24.98584 \times x + 25.98259 \times (21.01 - x)}{100}$.
$2431.2 = 1894.5783 + 24.98584x + 545.8942 - 25.98259x$.
$2431.2 = 2440.4725 - 0.99675x$.
$0.99675x = 9.2725$.
$x \approx 9.30 \%$.
Thus,the abundance of $_{12}^{25} Mg$ is $9.30 \%$ and the abundance of $_{12}^{26} Mg$ is $21.01 - 9.30 = 11.71 \%$.