= Solution
Use the <Aalen–Johansen estimator> of the <cumulative incidence function>, updating the overall <Kaplan–Meier estimator> for either event:
$$
\Delta\widehat F_B(t)=\widehat S(t-)\frac{d_B(t)}{Y(t)},\qquad \widehat S(t)=\widehat S(t-)\left[1-\frac{d_A(t)+d_B(t)}{Y(t)}\right].
$$
<Right censoring> changes subsequent <risk sets>, not the <survival function> by itself. Starting with the given estimates at $a_k$, the complete calculation is
$$
\begin{array}{c|c|c|c|c|c}
\text{event time}&Y(t)&\widehat S(t-)&\Delta\widehat F_B(t)&\widehat F_B(t)&\widehat S(t)\\\hline
a_{k+1}&10&0.40&0.40/10=0.04&0.34&0.36\\
a_{k+2}&9&0.36&0&0.34&0.32\\
a_{k+3}&8&0.32&0.32/8=0.04&0.38&0.28
\end{array}
$$
The middle event is of the competing type: it decreases overall survival while leaving the disease <cumulative incidence function> unchanged at that instant. The event-free intervals leave every estimate and <risk set> unchanged.
At the final time, both the event subject and the subject censored at that time are in the just-before <risk set>, so its denominator is eight. This is the usual event-before-censoring convention for recorded ties. After the event and <right censoring>, six subjects remain at risk. Consequently
$$
\boxed{\widehat F_B(a_{k+3})=0.30+0.04+0.04=0.38.}
$$
The nine numbered source items are the inputs to this single calculation, not nine further questions.
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