(1) $P\{Z \leqslant 0 \mid X < 2\}, P\{Z \leqslant 0\}$;
(2) $Z$ 的概率密度.

(1) 由 $X$ 与 $Y$ 相互独立, 得
$$\begin{gathered} P\{Z \leqslant 0 \mid X < 2\}=P\{Y-X \leqslant 0 \mid X=1\}=P\{Y \leqslant 1 \mid X=1\} \\ =P\{Y \leqslant 1\}=\int_{-\infty}^1 f_Y(y) \mathrm{d} y=\int_0^1 y \mathrm{~d} y=\frac{1}{2} . \\ P\{Z \leqslant 0\}=P\{Z \leqslant 0 \mid X=1\} P\{X=1\}+P\{Z \leqslant 0 \mid X=2\} P\{X=2\} \end{gathered}$$

\begin{aligned} =& \frac{1}{2} \times \frac{1}{3}+P\{Y \leqslant 2 \mid X=2\} P\{X=2\} \\ =& \frac{1}{2} \times \frac{1}{3}+P\{Y \leqslant 2\} P\{X=2\} \\ =& \frac{1}{2} \times \frac{1}{3}+1 \times \frac{2}{3}=\frac{5}{6} . \\ F_Z(z) &=P\{Z \leqslant z\}=P\{Y-X \leqslant z\} \\ &=P\{Y-X \leqslant z, X=1\}+P\{Y-X \leqslant z, X=2\} \\ &=P\{Y \leqslant z+1, X=1\}+P\{Y \leqslant z+2, X=2\} \\ &=P\{Y \leqslant z+1\} P\{X=1\}+P\{Y \leqslant z+2\} P\{X=2\} \\ &=\frac{1}{3} F_Y(z+1)+\frac{2}{3} F_Y(z+2), \end{aligned}

\begin{aligned} f_Z(z) &=F_Z^{\prime}(z)=\frac{1}{3} F_Y^{\prime}(z+1)+\frac{2}{3} F_Y^{\prime}(z+2)=\frac{1}{3} f_Y(z+1)+\frac{2}{3} f_Y(z+2) \\ &= \begin{cases}0, & z < -2, \\ 0+\frac{2}{3}(z+2)=\frac{2}{3}(z+2), & -2 \leqslant z < -1, \\ \frac{1}{3}(z+1)+\frac{2}{3}[2-(z+2)]=\frac{1}{3}(1-z), & -1 \leqslant z < 0, \\ \frac{1}{3}[2-(z+1)]+0=\frac{1}{3}(1-z), & 0 \leqslant z < 1, \\ 0, & z \geqslant 1\end{cases} \\ &= \begin{cases}\frac{2}{3}(z+2), \quad-2 \leqslant z < -1, & \\ \frac{1}{3}(1-z), & -1 \leqslant z < 1, \\ 0, & \text { 其他. }\end{cases} \end{aligned}