设函数f(x)=-1/3x^3+x^2+(m^2-1)x (x属于R),其中m大于0
（1）.当m=1时,求曲线y=f(x)在点（1,f(1))处切线的斜率.
f(x)=-1/3*x^3+x^2,
f'(x)=-x^2+2x,f'(1)=1,为所求.
（2）.求函数f(x)的单调区间和极值.
f'(x)=-x^2+2x+m^2-1
=-(x-1)^2+m^2
=-(x-1+m)(x-1-m),
由m>0知1-m