1) You'd use a t-test. In this case sample sizes are identical, so there's not even a need to worry about Welch adjustment; the ordinary t-test would do.
2) That depends; if you're prepared to assume identical distributions up to a location shift, then if population means are finite you could certainly use a Wilcoxon-Mann-Whitney for that. However, it's considerably more broadly applicable for a test of equality of means than that*. [One advantage of restriction to a location-shift alternative is interpretability, but it's not necessary for the test to be suitable. Scale shift alternatives - and a host of other transformations that imply ordered means - would all be included]
If you assume only identical distributions under the null (no need to restrict it to location shift alternative or even to a stochastic-dominance alternative), you could do a permutation test on the difference in means. (There are still other options though)
What software are you using?
* Here's an illustration that it can be broader and still be about means.
(i) Assume $X_i$, $Y_j$ are both continuous from $F_X$ and $F_Y$ respectively (each with finite mean), with the usual independence assumptions and assume for this argument that both variables are non-negative. Let $S_X=1-F_X$. Let's take our hypotheses to be
$H_0: F_X=F_Y$ vs $H_1: F_X<F_Y$
Then that's equivalent to
$H_0: S_X=S_Y$ vs $H_1: S_X>S_Y$
Integrate the expressions on both sides on the positive half line and we get:
$H_0: E(X)=E(Y)$ vs $H_1: E(X)>E(Y)$
as long as $\int_0^\infty S_X(t)-S_Y(t) dt >0$ and we can interchange the integration and the difference.
[This sort of argument can be easily extended to a situation with a lower bound on the variables other than zero.]
(ii) the restriction to positive random variables was a convenience in order to be able to bring in the argument about the relation between survival functions and expectation; it's not necessary. For example, second order stochastic dominance implies the expectations are ordered.
I'm not currently clear on what the weakest restrictions (on top of the assumptions already made for the WMW) above first order stochastic dominance would need to be to imply ordering of expectations, but it's whatever minimum assumption is needed for $F_X-F_Y<0$ in some region to imply $E(X)-E(Y)>0$. (I think if the inequality $F_X-F_Y\leq 0$ is somewhere strict in such a way that their integrals differ in that region, it should be possible to show that $E(X)-E(Y)>0$ follows.)
I hope to come back to this and say some more once I've thought about it (and maybe learned some more). The implication is that the Wilcoxon-Mann-Whitney is useful for testing means under much less restrictive assumptions than are often suggested.
As gung points out, we still need some assumptions. The Wilcoxon-Mann-Whitney is sensitive to an even broader set of alternatives than I'm discussing here, so without additional assumptions, rejection doesn't necessarily suggest a difference in means.