Skills of all levels.

Expanding $f(x)$ , we have:

$\begin{aligned} f(x) & = x + \frac 23 x^3 + \frac {2 \cdot 4}{3 \cdot 5}x^5 + \frac {2 \cdot 4 \cdot 6}{3 \cdot 5 \cdot 7}x^7 + \cdots \\ & = \sum_{k=0}^\infty \frac {(2k)!! x^{2k+1}}{(2k+1)!!} \\ & = \frac {\sin^{-1} x}{\sqrt{1-x^2}} & \small \color{#3D99F6} \text{Can prove it by Taylor series} \end{aligned}$

Therefore, $f \left(\sqrt{\frac {6057}{8076}}\right) = f \left(\frac {\sqrt 3}2\right) = \frac {\sin^{-1} \frac {\sqrt 3}2}{\sqrt{1-\frac 34}} = \frac 23 \pi$ and

$\begin{aligned} \left(\frac 6 \pi f \left(\sqrt{\frac {6057}{8076}}\right)\right)^{2019} & \equiv \left(\frac 6 \pi \cdot \frac 23 \pi \right)^{2019} \text{ (mod 2019)} \\ & \equiv 4^{2019} \text{ (mod 2019)} \\ & \equiv \boxed{64} \text{ (mod 2019)} & \small \color{#3D99F6} \text{See note.} \end{aligned}$

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Note: Note that $2019 = 3 \times 673$ has two prime factors. Consider each factor separately.

For factor 3: $4^{2019} \equiv (3+1)^{2019} \equiv 1^{2019} \equiv 1 \text{ (mod 3)}$

For factor 673: Since $\gcd (4, 673) = 1$ , Euler's theorem applies and since 673 is a prime, Euler's totient function $\phi (673) = 673-1=672$ . Then $4^{2019 \bmod \phi(673)} \equiv 4^{2019 \bmod 672} \equiv 4^3 \equiv 64 \text{ (mod 673)}$ .

Since $64 \equiv 1 \text{ (mod 3)}$ , $4^{2019} \equiv 64 \text{ (mod 2019)}$ .

@Chew-Seong Cheong , please post your solution.