(a) Expand \((2r+1)^3 - (2r-1)^3\):

\(8r^3 + 12r^2 + 6r + 1 - (8r^3 - 12r^2 + 6r - 1) = 24r^2 + 2\).

Sum both sides and use the method of differences:

\(24 \sum_{r=1}^{n} r^2 + 2n = (2n+1)^3 - 1\).

\(24 \sum_{r=1}^{n} r^2 = 8n^3 + 12n^2 + 4n = 4n(n+1)(2n+1)\).

Thus, \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\).

(b) Express \(S_{2n}\) in terms of sums:

\(S_{2n} = \sum_{r=1}^{2n} r^2 + 2 \sum_{r=1}^{n} (2r)^2 = \sum_{r=1}^{2n} r^2 + 8 \sum_{r=1}^{n} r^2\).

Apply \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\):

\(S_{2n} = \frac{1}{6}(2n)(2n+1)(4n+1) + \frac{8}{6}n(n+1)(2n+1)\).

\(S_{2n} = \frac{1}{3}n(2n+1)(4n+4+4) = \frac{1}{3}n(2n+1)(8n+5)\).

(c) Evaluate the limit:

\(\lim_{n \to \infty} \frac{S_{2n}}{n^3} = \frac{16}{3}\).