Let's consider the monthly contributions. 10 years ago she started paying $625 to the retirement fund. After another 5 years that initial payment would increase to $625*(1+0.0384/12)^(15*12). The second month accumulates $625*(1+0.0384/12)^(15*12-1), one month less than the initial payment. The third month's period of growth will be measured over 15*12-2=178 months. So it continues up to the last payment which will only accumulate a month's interest. The monthly rate is 0.0384/12=0.0032, so the growth factor is 1.0032^p where p is the number of months. Now we can write a series, S, that takes into account the accumulated no interest. Let's use a general expression, where A is the monthly contribution amount, r the monthly rate as a fraction, and p=12n where n is the number of years:
S=A(1+r)^p+A(1+r)^(p-1)+A(1+r)^(p-2)+...+A(1+r)^2+A(1+r)
S=A(1+r)((1+r)^(p-1)+...+1).
This equation contains a geometric progression which can be summed.
(1+r)S=A(1+r)((1+r)^p+(1+r)^(p-1)+...+(1+r)^2+(1+r)).
So (1+r)S-S=A(1+r)((1+r)^p-1); rS=A(1+r)((1+r)^p-1).
This formula can be applied to both parts of this question.
a. Same rate applies for the whole 15 years. 0.0032S=625*1.0032(1.0032^180-1)=625*1.0032*0.7773=487.3504, so S=$152,297.
b. 0.0032 applies for ten years: 0.0032S=625*1.0032*(1.0032^120-1)=292.963, so S=$91,550.94. Now start a new series with A=1000 and r=0.0772/12=0.0064333... or 0.0193/3 and p=60. 0.006433S=1000*1.006433(1.006433^60-1)=472.286, so S=$73,412.33. However, we're not through yet, because $91,550.94 continues to earn interest over 5 years at the new rate. This comes to 91550.94*1.006433^60=$134,512.78. The combined amount is this plus 73,412.33=$207,925.10 approx. (Approximate figures throughout.)