a)

$3^2+{11}^0+{10}^1=9+1+10=20$

$7+3\cdot 5^3=7+3\cdot 125=7+375=382$

$\left(5^2-13\right)\cdot 4^2=\left(25-13\right)\cdot 16=12\cdot 16=\left(10+2\right)\cdot 16=10\cdot 16+2\cdot 16=160+32=192$

b)

${\left(1\frac{1}{7}\right)}^2+\frac{4^3}{7}-{\left(\frac{5}{8}\right)}^0={\left(\frac{8}{7}\right)}^2+\frac{64}{7}-1=\frac{8^2}{7^2}+\frac{64}{7}-1=\frac{64}{49}+\frac{64\cdot 7}{7\cdot 7}-1=$