Let the first term of AP be a and its common difference be d.
So by first condition we get t3+t7=(a+2d)+(a+6d)= 6
=> a= 3-4d............[1]
By 2nd condition we get t3xt7=(a+2d)x(a+6d)= 8
=> (3-4d+2d)x(3-4d+6d)= 8
=> (3-2d)x(3+2d)= 8
=>9-4d2=8
=>d2=1/4
=> d=+0.5 and -0.5
When d = 0.5 then a= 1
and when d= -0.5 then a= 5
when a=1 and d =0.5 the sum of first 16 terms will be
S16= 16/2[2x1+(16-1)x0.5]=76
when a=5 and d =-0.5 the sum of first 16 terms will be
S'16= 16/2[2x5-(16-1)x0.5]=20