Use input that forces compatibility at the two junctures. In other places these functions are continuous.
solve a*(atan(0)+2)=3, 2*e^(3b)+1=9 for a, b
http://www.current.wolframalpha.com/input/?i=solve+a*%28atan%280%29%2B2%29%3D3%2C+2*e^%283b%29%2B1%3D9+for+a%2C+b
In Mathematica this could be done in any of several ways. Here is one such.
Solve[{a*(ArcTan[0] + 2) == 2*E^b + 1, 2*E^(3*b) + 1 == Log[1] + 3^2}, {a, b}, Reals]
This is more elaborate but allows one more flexibility in handling.
f1[a_, b_, x_] := a*(ArcTan[x] + 2)
f2[a_, b_, x_] := 2*Exp[b*x] + 1
f3[a_, b_, x_] := Log[x - 2] + x^2
Solve[{f1[a, b, 0] == f2[a, b, 0],
f2[a, b, 3] == f3[a, b, 3]}, {a, b}, Reals]
(* Out[14]= {{a -> 3/2, b -> (2 Log[2])/3}} *)