a)

Thus, the rate of decay of A is 0.00666 M/s and the rate of formation of B is 0.0133 M/s.

b)
There are two ways to do this problem. The hard way is to notice that the reaction is first order in A and thus A will follow the integrated rate law: A(t) = A0e-kt. Since A starts out at 0.3 M (=A0), then we have A(1 sec) = 0.3exp(-0.01 x 1)=0.293M. If A dropped in amount by 0.0066 M, B must have increased by 0.0132 M, giving B(1 sec)=0.0132 M, and C must have increased by 0.0198 M so that C(1 sec)=0.0198 M.
The easy way to do this problem is to recognize that 1 second is short on the timescale of this reaction ( a first order rate constant of 0.0222 per second means that the reaction takes about 45 seconds to become two thirds complete). Therefore we can use the differentials directly to solve the problem: dA = -0.00666dt = -0.00666 M so A(1 sec) = 0.3 - 0.00666 = 0.293 M. The rest is the same ast above.