The Poisson distribution $\lambda$ is $1.8$. The probability of no event in an hour is $e^{-\lambda}$.
Another way: The waiting time has exponential distribution with parameter $\lambda$. Now calculate as you did, with parameter $1.8$.
Say we call our two bulb lifetimes $X$ and $Y$ and assume they are independent and identically distributed.
That the distributions of the lifetimes of light bulbs are memoryless does not seem plausible to me, but I'll play along.
We have $\Pr(X>x) = e^{-x/1000} = \Pr(Y>x)$.
So
\begin{align}
& \Pr(\min\{X,Y\} > x) = \Pr(X>x\ \&\ Y>x) = \Pr(X>x)\cdot\Pr(Y>x) \\[10pt]
= {} & \Big( e^{-x/1000}\Big)^2 = e^{-x/500}.
\end{align}
This is an exponential distribution with expected value $500$, so that's how long it is on average until the first bulb burns out.
The average time until they have both burned out is $\operatorname{E}(\max\{X,Y\})$. We have
\begin{align}
& \Pr(\max < x) = \Pr(X<x\ \&\ Y<x) = \Pr(X<x) \cdot \Pr(Y<x) \\[10pt]
= {} & (1-e^{-x/1000})^2 = 1 - 2e^{-x/1000} + e^{-x/500}
\end{align}
So the density function is
$$
f_{\max} (x) = \frac d {dx} \left( 1 - 2e^{-x/1000} + e^{-x/500} \right) = 2\cdot\frac 1 {1000} e^{-x/1000} - \frac 1 {500} e^{-x/500} \text{ for } x>0.
$$
Since this density is $2$ times the density of an exponential distribution with mean $1000$ minus an exponential distribution with mean $500$, we get
\begin{align}
& \int_0^\infty x f_{\max}(x)\,dx \\[10pt]
= {} & 2\int_0^\infty x\cdot(\text{that first density})\,dx - \int_0^\infty x\cdot(\text{that second density}) \,dx \\[10pt]
= {} & 2\cdot1000 - 500 = 1500.
\end{align}
The answer to the third question is merely the sum of the two averages: $1000+1000=2000$.
Best Answer
Since the interarrival times are exponential random variables with rate $\lambda$, the probability for any given one of them to exceed $T$ is $e^{-\lambda T}$, and they are independent. So the number of arrivals $N$ needed for this to happen is a geometric random variable with parameter $p = e^{-\lambda T}$. Its mean is thus $1/p = e^{\lambda T}$.