If the probabilities of the three components $A$, $B$, $C$ were independent of each other (in other words: uncorrelated), then it would indeed be correct to simply multiply the probabilities associated with $ABC$, $AB$, $AC$ and $BC$ working. Followed by adding up these four numbers, giving you the chance that least two components work.
However, as clearly stated in the text, the probabilities for components $B$ and $C$ are interdependent! This means that, before you can do a calculation as in the previous paragraph, you must first explore fully the interdependence of these probabilities.
Okay, let us do this. It is given that $C$ has a failure chance of $0.01$. Furthermore it is given that if $C$ fails, then the chance that $B$ fails is $0.55%$. This leads us to the conclusion that the probability that both $C$ and $B$ fail is equal to $0.01 * 0.55 = 0.0055$.
With a bit more effort we can derive that the probability that $C$ fails and $B$ works is equal to $0.0045$. The chance that $C$ works and $B$ fails is $0.1455$. The chance that both $C$ and $B$ work is $0.8455$. We now have the four values that determine the interdependence of $C$ and $B$.
Finally we can compute the overall probability that the system works. We get:
$$P = P(ABC) + P(AC) + P(AB) + P(BC)$$
$$P = 0.7 * (0.8455 + 0.1445 + 0.0045) + 0.3 * 0.8455 = 0.9498$$
Best Answer
So this question is tricky because you are given the cdf (cumulative distribution function) which I will call $F(x)$, but in order to calculate expectation you need the pdf (probability distribution function) which I will call $f(x)$. We can find the pdf by noting that $$ F(x)=P\{X\leq x\}=1-P\{X>x\}=1-\frac{1}{(x+1)^3}=\int_0^xf(x)\,dx $$ Now, we take a closer look at that last bit of the equality: $$ 1-\frac{1}{(x+1)^3}=\int_0^xf(x)\,dx $$ By the fundamental theorem of calculus, we can differentiate both sides to find $$ f(x) = \frac3{(x+1)^4} $$ In order to evaluate the expected lifetime, we find $$ E(x)=\int_0^\infty x\,f(x)\,dx=\int_0^\infty \frac{3x}{(x+1)^4}\,dx $$ As you stated, the integral (and thus the expectation) should come out to $\frac12$.
That the other answerer's method should work equally well is explained here http://en.wikipedia.org/wiki/Expected_value#Continuous_distribution_taking_non-negative_values
As for the second part, the probability of a single component not failing in the first 2 months is $$ p=P(X\geq 2)=\frac1{(2+1)^3}=\frac1{27} $$ The probability that this happens exactly 2 out of 5 times is simply $$ P=\binom{5}{2}p^2(1-p)^3 $$