Hi Abinash,
Solving individual parts-

Part 1 : $$\frac{\left(2^{\frac{1}{2}}\times\ 3^{\frac{1}{3}}\times\ 4^{\frac{1}{4}}\right)}{10^{-\frac{1}{5}}\times\ 5^{\frac{3}{5}}}$$
We can write 4 as 2^2 and 10 = 5*2

Substituting these values here, we get
$$\frac{\left(2^{\frac{1}{2}}\times3^{\frac{1}{3}}\ \times\ 2^{\frac{2}{4}}\right)}{2^{-\frac{1}{5}}\times\ 5^{-\frac{1}{5}}\times\ 5^{\frac{3}{5}}}$$
Now, adding all these powers
=>$$\frac{\left(2^{\frac{1}{2}+\frac{2}{4}}\times3^{\frac{1}{3}}\ \right)}{2^{-\frac{1}{5}}\times\ 5^{-\frac{1}{5}+\frac{3}{5}}}$$

=>$$\left(2^{\frac{1}{2}+\frac{2}{4}+\frac{1}{5}}\times3^{\frac{1}{3}}\ \times\ 5^{\frac{1}{5}-\frac{3}{5}}\right)$$ (Taking powers from denominator to numerator, the sign changes)
=> $$\left(2^{\frac{6}{5}}\times3^{\frac{1}{3}}\ \times\ 5^{-\frac{2}{5}}\right)$$  -----(1)
Similarly, for equation 2
$$\frac{\left(3^{\frac{4}{3}}\times\ 5^{-\frac{7}{5}}\right)}{4^{-\frac{3}{5}}\times\ 6}$$
We can write 4 as 2^2 and 6 = 2*3
Substituting these values here, we get
$$\frac{\left(3^{\frac{4}{3}}\times\ 5^{-\frac{7}{5}}\right)}{2^{-\frac{6}{5}}\times\ 2\times\ 3}$$
=> $$\left(3^{\frac{1}{3}}\times\ 2^{\frac{1}{5}}\times\ 5^{-\frac{7}{5}}\right)$$  -----(2)
Now divide both 1 and 2 in a similar way, you will get the answer.
I hope this helps!