Let \(X\) be the random variable for the weight of any given can, and let \(\mu\) and \(\sigma\) be the mean and standard deviation, respectively, for the distribution of \(X\).
You have
\(\begin{cases}\mathbb P(X<347)=0.0015\\\mathbb P(X>377)=0.025\end{cases}\)
Recall that for any normal distribution, approximately 99.7% of it lies within three standard deviations of the mean, i.e. \(\mathbb P(\mu-3\sigma. This means 0.3% must lie outside this range, \(\mathbb P(X<\mu-3\sigma~\lor~X>\mu+3\sigma)=0.003\). Because the distribution is symmetric, it follows that \(\mathbb P(X<\mu-3\sigma)=0.0015\).
Also recall that for any normal distribution, about 95% of it falls within two standard deviations of the mean, so \(\mathbb P(\mu-2\sigma, which means 5% falls outside, and by symmetry, \(\mathbb P(X>\mu+2\sigma)=0.025\).
Together this means
\(\begin{cases}\mu-3\sigma=347\\\mu+2\sigma=377\end{cases}\)
Solving for the mean and standard deviation gives \(\mu=365\) and \(\sigma=6\).
