Studying for an FIA actuarial exam? 8-)
As @whuber said, knowing the expected value is key. What is an expected value? Simply (and probably non-rigorously), it can be thought of as the long-term average. If the company wrote this policy for one million blocks of three years, then the total amount they paid over those blocks divided by a million would be very close to the "expected value".
Insurance works on the principle that even though any one person's risk is too volatile to measure with any certainty, a pool of risks is much more tractable and measurable. So while said insurer would not want to write just one policy, if they write 100,000 such policies, they will have a much better chance of having the actual average loss they experience be close to the expected value. Because there is volatility, one must add a contingency/volatility/profit load which in this case is the $£100$.
Now, to the case in question - the expected value in this case would be the sum of the expected payments over the three years. There is a $10\%$ chance the system fails in year $1$. If it does, the payout is $£3000$ so the expected payment in year 1 is $£300$. For the system to fail in year $2$ it must have survived year $1$, so the expected loss is $0.9\cdot0.1\cdot£2000$ or $£180$. Similarly for year $3$ the system has to survive for two years and then fail, so the expected loss is $0.9^2\cdot0.1\cdot£1000$ or $£81$. So the total expected loss for the insurance policy is $£561$. What this means is if the company writes one million of these policies, they can be rather sure that their sum total loss will be very close to $£561,000,000$. Since the company wants to make an average profit of $£100$ one each policy, they should charge $£661$.
As for the second question, it is a matter of understanding the proportion that each ring and center bullseye is of the the total area of the circle, as having a uniform distribution means that the shooter is rather wild and can land the shot on any point within that circle equally. The area of the total circle is $64\pi$. The area of the bullseye is $1^2\pi$, the area of the five-point ring is the area of its circle minus the area of the bullseye, which is is $3^2\pi - 1^2\pi = 8\pi$. Similarly, the area of the three-point ring is $5^2\pi - 3^2\pi = 16\pi$ and there are no points in the remaining $8^2\pi - 5^2\pi = 39\pi$. The $\pi$ all cancel in the ratios, and we have the following probability:
$$
\frac{0\cdot 39 + 3 \cdot 16 + 5 \cdot 8 + 10 \cdot 1}{64} = \frac{98}{64} \approx 1.53
$$
So this persons expected "points-per-shot" is $\frac{98}{64}$ or about $1.53$.
