Why isn't there an $x$ such that $\tan\left(\frac{\sin^{-1} x}{5}\right)=1$? If $\tan\left(\frac{\sin^{-1} x}{5}\right)=1$, then $x$ is equal to 

$$\frac{-\pi}{2}\le \sin^{-1}x \le \frac{\pi}{2}$$
$$\frac{-\pi}{10}\le \frac{\sin^{-1}x}{5} \le \frac{\pi}{10}$$

This show that no value of such $x$ exist.
I want to know why this is happening.

Now in this approach,
$$\frac{\sin^{-1}x}{5}=\frac{\pi}{4}$$
$$\sin^{-1}x=\frac{5\pi}{4}$$
Why this is happening?

I think this is just happening because of principal value  of $\sin^{-1}x$.
Indeed, the range for $\arcsin(x)$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$, so no real $x$ satisfies $\arcsin(x)=\frac{5}{4}\pi$.
$\sin^{-1} x = \frac{5 \pi}{4}$ gives $x=\sin \frac{5 \pi}{4} = \sin (\pi + \frac{\pi}{4})$, then $x = -\sin \frac{\pi}{4} = -\frac{1}{\sqrt 2}$. Your error is restricting $\sin^{-1} x$ to the values $-\frac{\pi}{2} \le \sin^{-1} x \le \frac{\pi}{2}$ (the principal value is within this range).