Probability of lamp after $10500$ h when already reached $9000$ h
Solution 1:
If $X$ is the random variable that represents lifetime of the bulb,
$P(X \gt 9000) \ne 0.1056$. Rather $P(X \gt 9000) = 1 - 0.1056$. The integral will be,
$ \displaystyle P(X \gt 9000) = \int_{9000}^{\infty} \frac {1}{800 \sqrt{2 \pi}} e^{- (x - 10000)^2 / 1280000} ~dx$ $\approx 0.89435$
$P(X \gt 10500) \approx 0.266$ as you obtained.
So, $ \displaystyle P(X \gt 10500 | X \gt 9000) \approx \frac{0.266}{0.89435}$