I don't know how to do part (c) and (d)
For (c), you are looking for the probability that $\displaystyle 850<X<1350$, while eliminating the area of the graph up to $\displaystyle X=850$.
which is $\displaystyle \displaystyle\frac{P(X<1350)-P(X<850)}{P(X>850)}$
Convert to the Z-values and compute the probability.
Bear in mind that the probability of the bulb failing within the first 850 hours is now zero.
Hence your fraction has $\displaystyle P(X>850)$ in the denominator.
For (d), use the Binomial conversions $\displaystyle \sigma=\sqrt{npq},\;\;\; \mu=np$
where p=0.07, q=0.93, n=1200
Let P(t) be the CDF of the bulb life.
For c you want [P(1350)-P(500)]/(1-P(500)) Imagin you have 1200 bulbs then 1200 P(500) have failed by 500 hours, 1200 P(1350) by 1350 hours, so out of the 1200 (1-P(500)) that survive the first 500 hours 1200 [P(1350)-P(500)] fail in the next 850 hours.
CB