Demonstration that EXP is closed under union complementation and concatenation
By definition $EXP = \bigcup_{k\in N}{DTIME\big(2^{n^k}\big)}$. 

Let $L_1, L_2 \in EXP$. Then $L_1 \in DTIME\big(2^{n^m}\big)$ and $L_2 \in DTIME\big(2^{n^k}\big)$ for some $k,m \in N$. 

So for a string $w$ checking $w \in L_1 \cup L_2$ takes $2^{n^m} + 2^{n^k}$ time which is $O(2^{n^l})$ where $l = max(m,k)$. So $L_1 \cup L_2 \in DTIME\big(2^{n^l}\big)$ and hence in $EXP$.

Let $w$ be a string of length $t$. Then checking $w \in L_1L_2$ takes $t(2^{n^m} + 2^{n^k})$ since for strings $u$ and $v$ such that $w=uv$ we check $w \in L_1$ and $v \in L_2$ $t+1$ times since there are $t$ different prefixes and suffixes of $w$. Assuming $l = max(m,k)$ we have $t2^{n^l}$ running time which is less than $2^t2^{n^l} = 2^{t + n^l}$ which belongs to $EXP$.   

Finally, checking $w \in L_1$ takes $O(2^{n^m})$ times so the Turing machine halts with "no" if $w \in L_1$, with "yes" otherwise. Thus $\overline{L}_1^{} \in EXP$.