Example $\mathbf{5}$ 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
$$S \cdot (b+a) \geqslant (a+1+b+3)^{2},$$

Thus, $S \geqslant \frac{(a+b+4)^{2}}{a+b} = (a+b) + \frac{16}{a+b} + 8 \geqslant 2 \sqrt{16} + 8 = 16,$

Equality holds when $a=\frac{7}{3}, b=\frac{5}{3}$. Therefore, the minimum value of $S$ is 16.
2) By (8) (Hölder's inequality), we have
$$\left(\frac{(a+1)^{3}}{b^{2}} + \frac{(b+3)^{3}}{a^{2}}\right)(b+a)(b+a) \geqslant (a+1+b+3)^{3},$$
Thus,
$$\begin{aligned}
T & \geqslant \frac{(a+b+4)^{3}}{(a+b)^{2}} \\
& = x + 12 + \frac{48}{x} + \frac{64}{x^{2}} \quad (\text{let } a+b=x) \\
& = x + \frac{64}{x} + \left(\frac{64}{x^{2}} - \frac{16}{x} + 1\right) + 11 \\
& = \left(x + \frac{64}{x}\right) + \left(\frac{8}{x} - 1\right)^{2} + 11 \\
& \geqslant 2 \sqrt{64} + 0 + 11 = 27,
\end{aligned}$$

Equality holds when $a=\frac{22}{5}, b=\frac{18}{5}$. Therefore, the minimum value of $T$ is 27.