(1)a=2时,f(x)=lnx+1−x2x,(x>0),且f(1)=0,
又∵f(x)=2x−12x2,(x>0),
∴f(x)在x=1处的切线斜率为f′(1)=12,
故切线的斜率为y=12(x−1),
即x−2y−1=0;
(2)由题意,f′(x)=1x−1ax2=ax−1ax2,
∵a为大于零的常数,
若使函数f(x)在区间[1,+∞)上单调递增,
则使ax−1⩾0在区间[1,+∞)上恒成立,
即a−1⩾0,故a⩾1;
(3)①当a⩾1时,f(x)在区间[1,2]上单调递增,
则fmin(x)=f(1)=0;
②当0a⩽12时,f′(x)在区间[1,2]恒不大于0,
f(x)在区间[1,2]上单调递减,
则fmin(x)=f(2)=ln2−12a;
③当12a1时,令f′(x)=0可解得,x=1a∈(1,2);
易知f(x)在区间[1,1a]单调递减,在[1a,2]上单调递增,
则fmin(x)=f(1a)=ln1a+1−1a;
综上所述,
①当a⩾1时,fmin(x)=0;
②当12a1时,fmin(x)=ln1a+1−1a;
③当0a⩽12时,fmin(x)=ln2−12a.