(1)解析：∵f（x）=x^2+ax+2,x∈[-5,5]
當(dāng)a=-1時(shí),f（x）=x^2-x+2= (x-1/2)^2+7/4
∴當(dāng)x∈[-5,1/2)時(shí),函數(shù)f(x)單調(diào)減,x∈[1/2,5]時(shí),函數(shù)f(x)單調(diào)增.
(2)∵函數(shù)f（x）在[-5,5]上為增函數(shù)
f（x）=x^2+ax+2= (x+a/2)^2+(8-a^2)/4
函數(shù)對(duì)稱軸為x=-a/2
-a/2<=-5==>a>=10
∴a的取值范圍為[10,+∞）