For part (a), we determine $g(y)$ for any $y \ge 6$. Applying the functional equation we have:
$$g(g(y)) = \frac{3y}{y+3}.$$
Also, as $2 < \frac{3y}{y+3} < 3$ we can apply $g$ to each side and apply the second condition to get:
$$g(g(g(y))) = g\left(\frac{3y}{y+3}\right) = \frac{1}{2}\left(\frac{3y}{y+3} + 1\right) = \frac{4y+3}{2y+6}.$$
But also from the opening condition applied to $g(y)$:
$$g(g(g(y))) = \frac{3g(y)}{g(y) + 3}.$$
Equating the reciprocal of these last two expressions yields:
$$\frac{1}{3} + \frac{1}{g(y)} = \frac{2y+6}{4y+3}$$
whence:
$$g(y) = \frac{12y + 9}{2y + 15}.$$
Substituting $y = 2021$ gives:
$$g(2021) = \frac{12 \times 2021 + 9}{2 \times 2021 + 15}.$$

For part (b) we note from the second condition that $g(3) = 2$ and $g(2) = 3/2$. Playing around recursively, we have:
$$\begin{aligned} g\left(\frac{3}{2}\right) &= g(g(2)) = \frac{3 \times 2}{2 + 3} = \frac{6}{5} \\ g\left(\frac{6}{5}\right) &= g\left(g\left(\frac{3}{2}\right)\right) = \frac{3 \times 3/2}{3/2 + 3} = 1 \\ g(1) &= g\left(g\left(\frac{6}{5}\right)\right) = \frac{3 \times 6/5}{6/5 + 3} = \frac{6}{7}. \end{aligned}$$
We spot a pattern and conjecture that, for $n = 2, 3, 4, \dots$
$$g\left(\frac{6}{n}\right) = \frac{6}{n+1}.$$
The proof is by induction; the result holds by direct computation for $n = 2$ and $n = 3$. For general $n$, we assume the result for $n - 1$ and we have
$$g\left(\frac{6}{n}\right) = g\left(g\left(\frac{6}{n-1}\right)\right) = \frac{3 \times 6/(n-1)}{6/(n-1) + 3} = \frac{18}{6 + 3n - 3} = \frac{6}{n+1}.$$
The result is now proved. Substituting $n = 6 \times 2021 = 12126$ gives:
$$g\left(\frac{1}{2021}\right) = \frac{6}{12127}.$$