Prove preimage resistance property Let $ f : \{0,1\}^* \rightarrow \{0,1\}^n$ and $ g : \{0,1\}^* \rightarrow \{0,1\}^m$. $h$ is created by concatenate f and g: $ h : \{0,1\}^* \to \{0,1\}^{n+m}$ with $h(x) = f(x) || g(x)$. How do I prove that both $f$ and $g$ must be preimage resistant if $h$ is to be preimage resistant? Is the same true for 2nd-preimage resistance?
Firstly a hash function $H : \{0,1\}^* \to \{0,1\}^{k}$ is preimage resistant if 

Given any hash $y$ it is infeasible to find any message $x$ such that $H(x)=y$ .

Now for your case lets assume that one of the functions, say $g$, is not preimage resistance.
Now suppose we have a $h-$hash of unknown message $x$, we know, from the construction of $h$ , that the last m bits of our hash are equal to $g(x)$. 

Since $g$ is not preimage resistant we somehow are able to find the value of $x$, which obviously means that we have found the message $x$ that results in a certain $h$-hash