1. Given that $a, b, c$ are non-negative real numbers, and $S=a+2 b+3 c, T=a+b^{2}+c^{3}$.
(1) Find the minimum value of $T-S$;
(2) If $S=4$, find the maximum value of $T$.
(1) $T-S=b^{2}+c^{3}-2 b-3 c=(b-1)^{2}+c^{3}-3 c-1 \geqslant c^{3}-3 c-1$. Let $f(c)=c^{3}-3 c-1, c \geqslant 0$, then $f^{\prime}(c)=3 c^{2}-1$, so $f(c)$ is monotonically decreasing on $(0,1)$ and monotonically increasing on $(1,+\infty)$, thus $f(c)_{\min }=f(1)=-3$. Therefore, when $b=c=1$, the minimum value of $T-S$ is -3.
(2) $a+2 b+3 c=4 \Rightarrow a=4-2 b-3 c$, where $0 \leqslant b \leqslant 2,0 \leqslant c \leqslant \frac{4}{3}$, and $0 \leqslant 2 b+3 c \leqslant 4$. Then $T=b^{2}+c^{3}+4-2 b-3 c=c^{3}-3 c+b^{2}-2 b+4$. Let $g(c)=c^{3}-3 c, 0 \leqslant c \leqslant \frac{4}{3}$, then $g^{\prime}(c)=3 c^{2}-1$, so $g(c)$ is monotonically decreasing on $(0,1)$ and monotonically increasing on $\left(1, \frac{4}{3}\right)$, thus $g(c)_{\max }=\max \left\{g(0), g\left(\frac{4}{3}\right)\right\}=0$. Therefore, $T \leqslant b^{2}-2 b+4=(b-1)^{2}+3 \leqslant 4$, with equality when $a=c=0, b=2$ or $a=4, b=c=0$.

In summary, the maximum value of $T$ is 4.