This is a binomial probability problem. Let the probability of success be p=0.6, and therefore failure, q=0.4. Success is 'thumbtack lands up'.
(a) The probability that it lands up exactly twice in five trials is
(5;2)(0.6)2(0.4)3= 5! / 2!(5−2)! (0.6)^2(0.4)^3=144 / 625
(b) The probability that it lands at least twice is 1- (no. times landing once in five):
1−(5;1)(0.6)(0.4)^4=1−5! / 1!(5−1)!(0.6)(0.4)^4=577 / 625
(c) The mean in a binomial distribution is given by np, so here,
μ=np=5×0.6=3
and the variance by npq, so
var=npq=5×0.6×0.4=1.2