I agree with Michael Chernick's answer, but think that it can be made a little stronger. Ignore the 0.05 cutoff in most circumstances. It is only relevant to the Neyman-Pearson approach which is largely irrelevant to the inferential use of statistics in many areas of science.
Both tests indicate that your data contains moderate evidence against the null hypothesis. Consider that evidence in light of whatever you know about the system and the consequences that follow from decisions (or indecision) about the state of the real world. Argue a reasoned case and proceed in a manner that acknowledges the possibility of subsequent re-evaluation.
I explain more in this paper:
http://www.ncbi.nlm.nih.gov/pubmed/22394284
[Addendum added Nov 2019: I have a new reference that explains the issues in more detail https://arxiv.org/abs/1910.02042v1 ]
This is not a problem of the t-test, but of any test in which the power of the test depends on the sample size. This is called "overpowering". And yes, changing the test to Mann-Whitney will not help.
Therefore, apart from asking whether the results are statistically significant, you need to ask yourself whether the observed effect size is significant in the common sense of the word (i.e., meaningful). This requires more than statistical knowledge, but also your expertise in the field you are investigating.
In general, there are two ways you can look at the effect size. One way is to scale the difference between the means in your data by its standard deviation. Since standard deviation is in the same units as your means and describes the dispersion of your data, you can express the difference between your groups in terms of standard deviation. Also, when you estimate the variance / standard deviation in your data, it does not necessarily decrease with the number of samples (unlike standard deviation of the mean).
This is, for example, the reasoning behind Cohen's $d$:
$$d = \frac{ \bar{x}_1 - \bar{x}_2 }{ s}$$
...where $s$ is the square root of the pooled variance.
$$s = \sqrt{\frac{ s_1^2\cdot(n_1-1) + s_2^2\cdot(n_2 - 1) }{ N - 2 } }$$
(where $N=n_1+n_2$ and $s_1$ and $s_2$ are the standard deviations in group 1 and 2, respectively; that is, $s_1 = \sqrt{ \frac{\sum(x_i-\bar{x_1})^2 }{n_1 -1 }} $).
Another way of looking at the effect size -- and frankly, one that I personally prefer -- is to ask what part (percentage) of the variability in the data can be explained by the estimated effect. You can estimate the variance between and within the groups and see how they relate (this is actually what ANOVA is, and t-test is in principle a special case of ANOVA).This is the reasoning behind the coefficient of determination, $r^2$, and the related $\eta^2$ and $\omega^2$ stats. Now, in a t-test, $\eta^2$ can easily be calculated from the $t$ statistic itself:
$$\eta^2 = \frac{ t^2}{t^2 + n_1 + n_2 - 2 }$$
This value can be directly interpreted as "fraction of variance in the data which is explained by the difference between the groups". There are different rules of thumb to say what is a "large" and what is a "small" effect, but it all depends on your particular question. 1% of the variance explained can be laughable, or can be just enough.
Best Answer
Wikipedia appears to have your answers. Here's an excerpt from the example statement of results:
The Wilcoxon signed-rank test is appropriate for paired samples, whereas the Mann–Whitney test assumes independent samples. However, according to Field (2000), the Wilcoxon $W$ in your SPSS output is "a different version of this statistic, which can be converted into a Z score and can, therefore, be compared against critical values of the normal distribution." That explains your $z$ score too then!
FYI, Wikipedia adds that, for large samples, $U$ is approximately normally distributed. Given all these values, one can also calculate the effect size $η^2$, which in the case of Wikipedia's example is 0.319 (a calculator is implemented in section 11 here). However, this transformation of the test statistic depends on the approximate normality of $U$, so it might be inaccurate with ns = 8 (Fritz et al., 2012).
P.S. The Kruskal–Wallis test's results should not be interpreted as revealing differences between means except under special circumstances. See @Glen_b's answer to another question, "Difference Between ANOVA and Kruskal-Wallis test" for details.
References
Field, A. (2000). 3.1. Mann-Whitney test. Research Methods 1: SPSS for Windows part 3: Nonparametric tests. Retrieved from http://www.statisticshell.com/docs/nonparametric.pdf.
Fritz, C. O., Morris, P. E., & Richler, J. J. (2012). Effect size estimates: current use, calculations, and interpretation. Journal of Experimental Psychology: General, 141(1), 2–18. PDF available via ResearchGate.