For $A\subset [0,1]$ we denote $A^{(2)}:=\{x^2,x\in A\}$ and we put $\mathcal B:=\{A\subset [0,1], A^{(2)}\mbox{ is Borel measurable}\}$. Since for a subset $A$ we have $([0,1]\setminus A)^{(2)}=\{x^2,x\notin A\}=[0,1]\setminus A^{(2)}$ ($t\mapsto t^2$ is one-to-one) and for a sequence $\{A_n\}$ we have $\left(\bigcup_{n\in\mathbb N}A_n\right)^{(2)}=\bigcup_{n\in\mathbb N}A_n^{(2)}$, $\mathcal B$ is a $\sigma$-algebra. Since for $0\leq a<b\leq 1$ we have $[a,b]^{(2)}=[a^2,b^2]$, $\mathcal B$ contains the Borel $\sigma$-algebra of $[0,1]$ which gives that for all $B\in\mathcal B[0,1]$, $B^{(2)}\in\mathcal B[0,1]$.
The inequality $m(A^{(2)})\leq 2m(A)$ is true for each $A$ which is a disjoint union of intervals contains in $[0,1]$, and the set $\mathcal M:=\{A\subset[0,1], m(A^{(2)})\leq 2m(A)\}$ is a monotone class, and we can conclude using monotone class theorem.
For the second problem, we show that $A+B$ is sequentially compact. If $\{a_n\}\subset A$ and $\{b_n\}\subset B$ are two sequences then we can extract a subsequence $\{a_{n_k}\}$ of $\{a_n\}$ which is converging to $a\in A$ and from $\{b_{n_k}\}$ a subsequence $\{b_{\varphi(k)}\}$ which converges to $b$. So the sequence $\{a_{\varphi(k)}+b_{\varphi(k)}\}$ converges to $a+b$ and $A+B$ is compact.
For the inequality $m(A+B)\geq m(A)+m(B)$ we deal with the case both $A$ and $B$ are intervals, then $B$ an arbitrary compact subset of the real line and finally the general case, using the fact that
$$m(K)=\inf\left\{\sum_{k=0}^{+\infty}m(I_k), I_k \mbox{ open interval and }K\subset\bigcup_{k\in\mathbb N}I_k\right\}.$$
