For #1, the answer is certainly no, since continuity could easily fail at x=0. Ex: Define f by f(0)=0, and f(x)=1+arctan(x) for x>0. Then f satisfies all those conditions, but isn't continuous at x=0.
However, it looks to be necessarily continuous at every point except at x=0:
From bounded and non-decreasing, the left and right limits for f exist at each x>0. Fix a>0. Suppose f(x)->m1 as x->a-, and f(x)->m2 as x->a+. Since f non-decreasing, must have m1<=f(a)<=m2. Let x1, x2, such that 0<x1<a and x2>a. Since f concave, for x1<=x<=x2, have f(x) >= the line through (x1,f(x1)) & (x2,f(x2)) evaluated at that x. Since f(x2)>=m2, the line through (x1,f(x1)) & (x2,f(x2)) evaluated at that x is >= the line through (x1,f(x1)) & (x2,m2) evaluated at that x, since those lines intersect at (x1,f(x1)) and the first line has the greater slope (one could work it out easily enough algebraically).
Thus, for x1<=x<=x2, have f(x) >= [ (m2-f(x1)) / (x2-x1) ] * (x-x1) + f(x1). The that (closed) inequality holds as take the limit x->a- (since we know the limit x->a- of f(x) exists and = m1). Get m1 >= [ (m2-f(x1)) / (x2-x1) ] * (a-x1) + f(x1). Since that (closed) inequality holds for all x2>a, it holds when take the (existing) limit as x2->a+.
Get: m1 >= [ (m2-f(x1)) / (a-x1) ] * (a-x1) + f(x1) = m2.
Therefore m1<=f(a)<=m2 and m2<=m1. Therefore m1=f(a)=m2. Thus both one-sided limits of f exist at x=a, and are equal (m1=m2), so the full limit exists at x=a and equals that value (which is not only m1 and m2, but also f(a)). Thus, as x->a, f(x)->f(a). Thus f is continuous at x=a.