$解:由题意,得a+b=0,c=\frac{1}{4},|d|=1,所以d=±1$
$(1)3a+3b-4c=3(a+b)-4c=3×0-4×\frac{1}{4}=0-1=-1$
$(2)当c=\frac{1}{4},d=1时$
$8c-d+cd=8×\frac{1}{4}-1+\frac{1}{4}×1=2-1+\frac{1}{4}=\frac{5}{4}$
$当c=\frac{1}{4},d=-1时$
$8c-d+cd=8×\frac{1}{4}-(-1)+\frac{1}{4}×(-1)=2+1-\frac{1}{4}=\frac{11}{4}\ $
$综上,c-d+cd的值为\frac{5}{4}或\frac{11}{4}$