$$ { 解：圆柱面x^2+y^2=a^2在M_0处的法向量为：} \\ { 令f=x^2+y^2-a^2，\boldsymbol{n}=\left( f_x,f_y,f_z \right) =\left( 2x,2y,0 \right) =\left( x,y,0 \right) ；} \\ { 于是：\boldsymbol{n}_{M_0}=\boldsymbol{n}_1=\left( x_0,y_0,0 \right) .} \\ { 曲面bz=xy在在M_0处的法向量为：} \\ { 令g=xy-bz，\boldsymbol{n}=\left( g_x,g_y,g_z \right) =\left( y,x,-b \right) ；} \\ { 于是：\boldsymbol{n}_{M_0}=\boldsymbol{n}_2=\left( y_0,x_0,-b \right) .} \\ { 两平面在M_0处法向量的夹角余弦为：} \\ { \cos \left< \left. \boldsymbol{n}_1,\boldsymbol{n}_2 \right> \right. \begin{array}{l} =\frac{\boldsymbol{n}_1\cdot \boldsymbol{n}_2}{\left| \boldsymbol{n}_1 \right|\cdot \left| \boldsymbol{n}_2 \right|}=\frac{\left( x_0,y_0,0 \right) \cdot \left( y_0,x_0,-b \right)}{\sqrt{x_{0}^{2}+y_{0}^{2}+0^2}\cdot \sqrt{y_{0}^{2}+x_{0}^{2}+\left( -b \right) ^2}}\\ =\frac{x_0y_0+y_0x_0}{\sqrt{a^2}\cdot \sqrt{a^2+b^2}}=\frac{2bz_0}{a\sqrt{a^2+b^2}}；\\ \end{array}} \\ { 于是夹角\left< \left. \boldsymbol{n}_1,\boldsymbol{n}_2 \right> \right. =\mathrm{arc}\cos \left( \frac{2bz_0}{a\sqrt{a^2+b^2}} \right) .} \\ $$