Triangle inequality and the square root of a metric space 
for (i) I know that the square root part is true but I don't know how to put it into words to prove it. 
For (ii) I just don't know how top apply the requirements for a metric space to the square root of another metric space. Just kind of confusing me

The prove it's a metric you need to show that 
0) $\sqrt{\rho(x,y)}$ is a non-negative real value function.
1) $\sqrt{\rho(x,y)} = 0 \iff x = y$
2)  $\sqrt{\rho(x,y)} =  \sqrt{\rho(y,x)}$ for all $x, y \in M$.
3)  $\sqrt{\rho(x,z)} \le \sqrt{\rho(x,y)} + \sqrt{\rho(y, z)}$ for all $x, y, z \in M$.
0) As $\rho(x,y)$ is a non-negative real function (as it is a metric) and the square root of any non-negative real number is also a non-negative real number.  So $\sqrt{\rho(x,z)} $ is a non-negative real function.
1) $\sqrt{v} = 0 \iff v = 0$.  $\rho(x,y) = 0 \iff x = y$.  Therefore $\sqrt{\rho(x,y)} = 0 \iff \rho(x,y) = 0 \iff x = y$.
2 and 3 I leave to you.
