(a) Mass of $_{3}^{6} L i$ lithium isotope, $m _{1}=6.01512 u$

Mass of $_{3}^{7} L i$ lithium isotope, $m =7.01600 u$

Abundance of $_{3}^{6} L i, n_{1}=7.5 \%$

Abundance of $_{3}^{7} L i, n_{2}=92.5 \%$

The atomic mass of lithium atom is given as:

$m=\frac{m_{1} n_{1}+m_{2} n_{2}}{n_{1}+n_{2}}$

$=\frac{6.01512 \times 7.5+7.01600 \times 92.5}{92.5+7.5}$

$=6.940934 u$

(b) Mass of $_{5}^{10} B$ boron isotope $_{5}^{10} B, m =10.01294 u$

Mass of boron isotope $_{5}^{11} B, m =11.00931 u$

Abundance of $_{5}^{10} B, n_{1}=x \%$

Abundance of $_{5}^{11} B, n_{2}=(100-x) \%$

Atomic mass of boron, $m =10.811 u$

The atomic mass of boron atom is given as

$m=\frac{m_{1} n_{1}+m_{2} n_{2}}{n_{1}+n_{2}}$ $10.811$$=\frac{10.01294 \times x+11.00931 \times(100-x)}{x+100-x}$

$1081.11=10.01294 x+1100.931-11.00931 x$

$\therefore x=\frac{19.821}{0.99637}=19.89 \%$

And $100-x=80.11 \%$

Hence, the abundance of is $_{5}^{10} B 19.89 \%$ and that of $_{5}^{11} B$ is $80.11 \%$