Short answer to first question is by using inner regularity of Lebesgue measure. That’s what you have done.
Note that $U$\ $C$ $\subseteq$ $O$ \ $C$
Hence by monotonicity of measure $\lambda_n$($U$ \ $C$) $< \epsilon$ which gives $\lambda_n(C)> 1-\epsilon$.
Now for the second question,
Complement (wrt open unit cube) of any compact set in open unit cube is open and Lebesgue measure of open set is always non-zero. These observations prove second question using additivity of measure.
As a prelude, let me first say that $A\times B$ is clearly in the product $\sigma$ algebra, since the product sigma algebra $\mathcal{F}\otimes \mathcal{G}$ of two measurable spaces is the sigma algebra generated by measurable 'rectangles'. i.e:
$\mathcal{F}\otimes \mathcal{G}= \sigma\Big( \{ E\times F: E\in \mathcal{F}, F\in \mathcal{G} \} \Big)$
Going by this definition, your set $A\times B$ is Lebesgue measurable in $\mathbb{R}^{n+m}$. However going by your criterion, recall first that for $A\in \mathcal{L}^n$ and $B\in \mathcal{L}^m$, we have:
$\lambda_n \otimes \lambda_m (A\times B)= \lambda_n(A)\times \lambda_m(B)$
Since $A,B$ are Lebesgue measurable, there exist $U_1,U_2$ open and $V_1,V_2$ closed such that:
$U_1\supseteq A \supseteq V_1$ , $U_2\supseteq B \supseteq V_2$ while $\lambda_n(U_1\setminus V_1)\leq \frac{1}{2} \tilde{\epsilon}$ and $\lambda_m(U_2\setminus V_2)\leq \frac{1}{2} \tilde{\epsilon}$
Notice that $U_1\times U_2\supseteq A\times B \supseteq V_1\times V_2$ while $U_1\times U_2$ is open and $V_1 \times V_2$ is closed. Since you can write:
$U_1\times U_2= \Big( U_1\times V_2 \Big) \sqcup \Big( U_1 \times (U_2 \setminus V_2) \Big)= \Big( V_1\times V_2 \Big) \sqcup \Big( (U_1\setminus V_1)\times V_2 \Big) \sqcup \Big( U_1 \times (U_2\setminus V_2) \Big) $
Then:
$\lambda_{n+m}\Big( (U_1\times U_2) \setminus (V_1 \times V_2) \Big)= \lambda_{n+m}\Big( (U_1\setminus V_1)\times V_2
\Big) + \lambda_{n+m}\Big( U_1 \times (U_2\setminus V_2)
\Big)= $
$= \lambda_n(U_1 \setminus V_1)\cdot \lambda_m(V_2)+ \lambda_n(U)\cdot \lambda_m(U_2 \setminus V_2)$
If you assume $\lambda_n(U_1), \lambda_m(U_2)\leq M$, then for $\tilde{\epsilon}= \dfrac{\epsilon}{2M}$, you've shown that the difference is of measure less than $\epsilon$. Otherwise, there is a standard argument (which I will elaborate if you request) of going from the finite measured case to the $\sigma$-finite case.
Best Answer
If $K$ is bounded there exists a number $M$ with $K \subset B(0,M)$. Since $B(0,M) \subset [-M,M]^n$ you get $$K \subset [-M,M]^n.$$ You don't need to know the Lebesgue measure of the cube. Rather, since the cube covers the bounded set $K$ the definition of the Lebesgue (outer) measure tells you that $$\lambda_n(K) \le Vol_n([-M,M]^n) = (2M)^n.$$
Another way to prove the other result is by contradiction. Suppose that $U$ is an open set and that $\lambda_n(U) = 0$. Since $U$ is open it contains a ball $B(x,r)$ so by monotonicity $\lambda_n(B(x,r)) = 0$ too.
If you don't know the Lebesgue measure of a ball this might not be enough to convince you of a contradiction, but using the translation invariance of the Lebesgue measure it implies $\lambda_n(B(0,r)) = 0$, and using the scaling property of the Lebesgue measure it implies in turn that $\lambda_n(B(0,2^k r)) = 2^k \lambda_n(B(0,r)) = 0$ for all $k$. These balls cover $\mathbb R^n$, so subadditivity gives you $$\lambda_n(\mathbb R^n) \le \sum_k \lambda_n(B(0,2^k r)) = 0$$ which should be apparently false.