We need the remainder when $(2021)^{3762}$ is divided by $17$.

First reduce the base modulo $17$: $2021 \div 17$ Since $17 \times 118 = 2006,$ we get $2021 - 2006 = 15.$ So, $2021 \equiv 15 \pmod{17}.$ Hence, $2021^{3762} \equiv 15^{3762} \pmod{17}.$

Now write $15 \equiv -2 \pmod{17}$. Therefore, $15^{3762} \equiv (-2)^{3762} = 2^{3762} \pmod{17}$ because $3762$ is even.

Use Fermat's little theorem: Since $17$ is prime and $\gcd(2,17)=1$, $2^{16} \equiv 1 \pmod{17}.$ Now reduce the exponent modulo $16$: $3762 = 16 \cdot 235 + 2.$ So, $2^{3762} = (2^{16})^{235} \cdot 2^2 \equiv 1^{235} \cdot 4 \equiv 4 \pmod{17}.$

Therefore, the remainder is $\boxed{4}.$