Let $n_1,n_2$ be the number of moles of gases $1$ and $2$.
We have $\dfrac{C_{p_1}}{C_{v_1}}=\dfrac{5}{3}$ and We know $C_{p_1}=C_{v_1}+R$
$\therefore \dfrac{5C_{v_1}}{3}=C_{v_1}+R,\dfrac{2C_{v_1}}{3}R_1,C_{v_1}=\dfrac{3R}{2}$
$A$ and $C_{p_1}=C_{ v_ 1 }+R=\dfrac{5R}{2}$
Also, $\dfrac{C_{ p_ 2}}{C_{ v_ 2 }}=\dfrac{4}{3}$ and
$C_{p_2}=C_{v_2}+R$
$\therefore \dfrac{4C_{ v_ 2 }}{3}=C_{ v_ 2 }+R, \dfrac{C_{ v_ 2 }}{3}=R,C_{ v_ 2 }=3R$
$C_{ p_ 2}=C_{ v_ 2 }+ R=4R$
Now $C_{ v_ {mix} }=\dfrac{n_1C_{ v_ 1 }+n_2C_{ v_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{3R}{2}+3\times 3R}{2+3}$
$=\dfrac{3R+9R}{5}=\dfrac{12R}{5}$
$C_{ p_ {mix} }=\dfrac{n_1C_{ p_ 1 }+n_2C_{ p_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{5R}{2}+3\times 4R}{2+3}$
$=\dfrac{17R}{5}$
$l_{mix}=\dfrac{C_{ p_ {mix} }}{C_{ v_ {mix} }}=\dfrac{17}{12}=1.4166=1.42$.