Let [tex]X[/tex] be the random variable for the weight of any given can, and let [tex]\mu[/tex] and [tex]\sigma[/tex] be the mean and standard deviation, respectively, for the distribution of [tex]X[/tex].
You have
[tex]\begin{cases}\mathbb P(X<347)=0.0015\\\mathbb P(X>377)=0.025\end{cases}[/tex]
Recall that for any normal distribution, approximately 99.7% of it lies within three standard deviations of the mean, i.e. [tex]\mathbb P(\mu-3\sigma<X<\mu+3\sigma)=0.997[/tex]. This means 0.3% must lie outside this range, [tex]\mathbb P(X<\mu-3\sigma~\lor~X>\mu+3\sigma)=0.003[/tex]. Because the distribution is symmetric, it follows that [tex]\mathbb P(X<\mu-3\sigma)=0.0015[/tex].
Also recall that for any normal distribution, about 95% of it falls within two standard deviations of the mean, so [tex]\mathbb P(\mu-2\sigma<X<\mu+2\sigma)=0.95[/tex], which means 5% falls outside, and by symmetry, [tex]\mathbb P(X>\mu+2\sigma)=0.025[/tex].
Together this means
[tex]\begin{cases}\mu-3\sigma=347\\\mu+2\sigma=377\end{cases}[/tex]
Solving for the mean and standard deviation gives [tex]\mu=365[/tex] and [tex]\sigma=6[/tex].
