This question posed quite some problems for a good number of candidates who attempted it.  In the part (a), some candidates could not recognise the problem, as an A.P. problem.  A good number of them counted these numbers out and thereby wasted time. They were expected to recognize the first number divisible by 7 as the first term of the A.P. (i.e. 77) and the last number (497) as the last term and substitute these values in the formula Tn = a+ (n-1)d (where Tn = value of the nth term, a = first term, d = common difference and n = required term) i.e. a = 77, d = 7, Tn = 497 and after a little computation yields n, the required number to be equal to 61.
In part (b) some candidates were able to compare the 3rd term with the 8th term to obtain the equation a + 7d = 5(a + 2d) but could not get the second equation a + 6d = a + 3d + 9.  Some others were able to obtain  d = 3 by solving the 2nd equation correctly but could not obtain the value of a nor the required 5 terms.  From the second equation, bringing the like terms together gave 3d = 9. Hence, d = 3.Similarly, from the first equation, 4a = - 3d. Substituting 3 for d in the equation gave 4a = -3(3) = -9. Therefore, a = - 9/4.  The first .five terms of the AP were -9/4, ¾, 15/4, 22/4, 39/4.