I want to find the mean and variance of the third arrival time of a Poisson process in terms of years.
First, 1 event every 5 years occurs, which implies $\lambda=0.2$. Second, define $t_{k}$ the arrival time corresponding to the $k$-th event with $k=1,2,3,...$ and $\tau_{k}$ the i.i.d. exponential inter-arrival times with $\tau_{k}=t_{k}-t_{k-1}$ and $t_{0}=0$. These definitions imply the time of the third event, $t_{3}$, to be defined as $t_{3}=\tau_{1}+\tau_{2}+\tau_{3}=(t_{1}-t_{0})+(t_{2}-t_{1})+(t_{3}-t_{2})=t_{3}$.
Since the inter-arrival times are i.i.d. random variables (as claimed by the question itself and supported by the fact that events or increments governed by a Poisson process are independant), I concluded
$$\mathbb{E}[t_{3}]=\mathbb{E}[\tau_{1}+\tau_{2}+\tau_{3}]=\mathbb{E}[\tau_{1}]+\mathbb{E}[\tau_{2}]+\mathbb{E}[\tau_{3}]=\frac{1}{\lambda}+\frac{1}{\lambda}+\frac{1}{\lambda}=\frac{3}{\lambda}$$
and due to independance of the inter-arrival times
$$2\mathbb{C}ov[\tau_{1}+\tau_{2},\tau_{3}]=0=2\mathbb{C}ov[\tau_{1},\tau_{2}]$$
implying
$$\mathbb{V}ar[t_{3}]=\mathbb{V}ar[\tau_{1}+\tau_{2}+\tau_{3}]=\mathbb{V}ar[\tau_{1}]+\mathbb{V}ar[\tau_{2}]+\mathbb{V}ar[\tau_{3}]=\frac{1}{\lambda^{2}}+\frac{1}{\lambda^{2}}+\frac{1}{\lambda^{2}}=\frac{3}{\lambda^{2}}$$
This reasoning seems blunt to me and I have tried to find any form confirmation, but did not succeed, therefore could anyone provide an explanation why this answer is right/wrong?
Thank you in advance.
Your approach and result are correct, except for one detail: You unnecessarily invoked independence in order to apply the linearity of expectation. The linearity of expectation does not require independence (which is what makes it such a powerful tool). You would have $\mathbb E[\tau_1+\tau_2+\tau_3]=\mathbb E[\tau_1]+\mathbb E[\tau_2]+\mathbb E[\tau_3]$ even if the $\tau_i$ were not independent. Since the variance can be written as a difference of expectations, $\mathbb V\mathsf{ar}[X]=\mathbb E\left[X^2\right]-\mathbb E[X]^2$, the variance, too, can be calculated using the linearity of expectation without requiring independence. However, this would involve evaluating the expectations $\mathbb E[\tau_i\tau_j]$ for $i\ne j$, and you rightly avoided this by directly adding the variances (which does require independence).