Example 28 Let $a, b \in \mathbf{R}^{+}$.
(1) Find the minimum value of $S=\frac{(a+1)^{2}}{b}+\frac{(b+3)^{2}}{a}$;
(2) Find the minimum value of $T=\frac{(a+1)^{3}}{b^{2}}+\frac{(b+3)^{3}}{a^{2}}$.
(1) By Cauchy-Schwarz inequality, we have

So
$$\begin{array}{l}
S \cdot (b+a) \geqslant (a+1+b+3)^{2} \\
S \geqslant \frac{(a+b+4)^{2}}{a+b}=(a+b)+\frac{16}{a+b}+8 \\
\quad \geqslant 2 \sqrt{16}+8=16
\end{array}$$

When $a=\frac{7}{3}, b=\frac{5}{3}$, equality holds.
Thus, the minimum value of $S$ is 16.
(2) By (3) (Hölder's inequality), we have
$$\left(\frac{(a+1)^{3}}{b^{2}}+\frac{(b+3)^{3}}{a^{2}}\right)(b+a)(b+a) \geqslant (a+1+b+3)^{3}$$

So
$$\begin{aligned}
T & \geqslant \frac{(a+b+4)^{3}}{(a+b)^{2}}=x+12+\frac{48}{x}+\frac{64}{x^{2}} \, (\text{let } a+b=x) \\
& =x+\frac{64}{x}+\left(\frac{64}{x^{2}}-\frac{16}{x}+1\right)+11 \\
& =\left(x+\frac{64}{x}\right)+\left(\frac{8}{x}-1\right)^{2}+11 \geqslant 2 \sqrt{64}+0+11 \\
& =27 .
\end{aligned}$$

When $a=\frac{22}{5}, b=\frac{18}{5}$, equality holds.
Thus, the minimum value of $T$ is 27.