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a. for value of k, it seems its a continuous PDF, so we know one thing about PDF as,

$\int_{0}^{infinite} PDF = 1$

but here PDF is divided in intervals,so

$\int_{0}^{6}kx2 + \int_{6}^{12} k(12-x)^{2} = 1$

solve it you will get k = $\frac{1}{144}$

b.  probability of storm duration > 8hr

in this case our range would be    8 <= x <= 12

p = $\int_{8}^{12} \frac{1}{144} (12-x)^{2}$

solving this gives, p = 4/27....(i m not sure, please cross check)

c. in this case probability we have to find in interval 5 to 7 which is divided in two interval by 6

so our 2 intervals would be  5 to 6, and 6 to 7

p = $\int_{5}^{6} \frac{1}{144} x^{2} + \int_{6}^{7} \frac{1}{144} (12-x)^{2}$

solve it you will get, p = $\frac{91}{ 3 * 144}$ ..... please cross check
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This solves the first part of ques. I hope u can solve the second one too.