如图

1,
f(m+n) = f(m) + f(n) -1,
x>0, f(x) >1,

m = n = 0;
f(0) = 2f(0) - 1,
f(0) = 1,
x>0, f(x) > f(0) = 1;
m = -n,
f(0) = f(m) + f(-m) -1;
f(-m) = 2 -f(m);

x2 > x1,
x2 - x1 > 0,
f(x2-x1) >1,
f(x2-x1) = f(x2) + f(-x1)-1 >1,
f(x2) + f(-x1) > 2;
f(x2) + 2-f(x1) >2;
f(x2)-f(x1) > 0,
f(x2) > f(x1),

f(x)在R上是增函数;

2,
f(3) = 4,
f(3) = f(1) + f(2) -1;
f(2) = f(1) + f(1) -1;
4 = 3f(1) -2;
f(1) = 2;

f(x)是R上的增函数,
f(a^2 + a-5) < 2;
f(a^2+a-5) <2 = f(1);

a^2+a-5 < 1;
a^2 + a -6 <0;
-3 < a< 2;

不等式的解集是：
-3 < a < 2。