i) no, $-u$ is conjugate to $v$. (Easy to see cause $i(u+iv) = -(v-iu)$)
ii) this is true only on simply connected domains. If $u$ is harmonic then the (locally unique) conjugate $v$ such that for one given point $p$, say $v(p) = c$, a prescribed constant, may be obtained locally as $v(z) -c = \int_0^1 \frac{d}{dt}v(p+t(z-p)) dt $ since, using the Cauchy Riemann equations ($u_x = v_y$ and $u_y = - v_x$), you can express this integral using the derivatives of $u$. (Instead of a straight line you can use any smooth curve, of course). Once $v$ is known in a neighbourhood it is uniquely determined along any smooth curve on which $u$ is defined (e.g. by an integral as above or by analytic continuation), but the values obtained may depend on the curve chosen. It can be shown that in simply connected domains the path chosen is not relevant, but on, say, an annulus, this is no longer true.
As for iii) I have to admit I don't know, but in general I'd expect the answer to be no. I'd start searching by looking at domains like \{(x,y): x>0, f_1(x) < y < f_2(x) \} where $f_1(0) = f_2(0) = f_1'(0) = f_2'(0) = 0$ (so you have a cusp) and then by looking at harmonic $u$ on such domains such that $u_x$ remains bounded as $x\rightarrow 0$ but $u_y\rightarrow \infty$. If you can find such $u$ the above mentioned definition of $v$ will be unbounded near $0$.