（1）直线$L$的方向向量$\overset{\rightarrow}{s} = \left\{1 , 4 , - 1\right\}$，平面$\pi$的法向量$\overset{\rightarrow}{n} = \left\{2 , 2 , 1\right\}$，
设$L$与$\pi$的夹角为$\theta \left(0 \leq \theta \leq \frac{\pi}{2}\right)$，
则$\sin \theta = \frac{\left|\overset{\rightarrow}{s} \cdot \overset{\rightarrow}{n}\right|}{\left|\overset{\rightarrow}{s}\right| \left|\overset{\rightarrow}{n}\right|} = \frac{\left|1 \times 2 + 4 \times 2 + \left(- 1\right) \times 1\right|}{\cdot} = \frac{9}{\cdot} = \frac{}{2}$，所以$\theta = \frac{\pi}{4}$；
（2）设$\frac{x - 1}{1} = \frac{y}{4} = \frac{z + 1}{- 1} = t$，则直线$L$的参数式方程为$\left\{\right. x = t + 1 , \\ y = 4 t , \\ z = - t - 1 ,$
将其代入平面方程得$2 \left(t + 1\right) + 2 \times 4 t - t - 1 - 1 = 0$，解得$t = 0$，故交点为$\left(1 , 0 , - 1\right)$.
又所求平面与$L$垂直，则平面的法向量可取为$\left\{1 , 4 , - 1\right\}$，
故所求平面方程为$x - 1 + 4 y - \left(z + 1\right) = 0$，即$x + 4 y - z - 2 = 0$.