(1)由题意可知：f(1/2)=f(1/4+1/4)=f(1/4)+f(1/4)-1=2
得f(1/4)=1.5.
f(1/4)=f(1/2-1/4)=f(1/2)+f(-1/4)-1
f(-1/4)=f(1/4)-f(1/2)+1=0.5
f(-1/2)=f(-1/4-1/4)=f(-1/4)+f(-1/4)-1=0
(2)在定义域R中,设a>b,令a=b+c,此时必定有c>0.由题意可知
f(a)=f(b+c)=f(b)+f(c)-1
则f(a)-f(b)=f(c)-1
又f(c)=f(c-1/2+1/2)=f(c-1/2)+f(1/2)-1=f(c-1/2)+1
因c>0.故c-1/2>-1/2.即有f(c-1/2)>0
所以
f(a)-f(b)=f(c-1/2)+1-1=f(c-1/2)>0
即在x的定义域内,当a>b时,恒有f(a)>f(b)
所以f(x)为单调递增那个第二问是不是有点麻烦了，我没看懂啊、额,第二问你静下心来慢慢看,就能明白,如果还有问题,再追问,说明下哪一步没看明白