(a): $P[X < 355] = P[Z < \frac{355 - 360}{4}] = P[Z < -1.25] = 1 - \Phi[1.25] = .1056$.
Part (a) is simple, but I included it because I was not sure if I should somehow use it to solve (b).
(b): $P[X < 355] = 0.025$ implies that $P[Z < \frac{355 - \mu}{4}] = \Phi[\frac{355 - \mu}{4}] = 0.025.$ Since the standard normal table starts at $\Phi[0] = .5$, this tells me I need to use symmetry to solve for $\mu$.
Then $P[X \geq 355] = P[Z \geq \frac{355 - \mu}{4}] = 0.975$. Now $0.975$ corresponds with $P[Z \leq 1.96] = 0.975$, but the problem is that we have $P[Z\geq \frac{355 - \mu}{4}]$.
The only thing I could think of from here is that $P[Z \geq \frac{355 - \mu}{4}] = P[Z \leq 1.96]$.
Is there a trick involving symmetry that I am not seeing?
By the symmetry of the standard normal about $0$, we want $\frac{355-\mu}{4}\approx-1.96$.