Last year ever of doing this and looks like I'm starting on a low. Haven't yet figured out how to fully solve Q10 lol. All I know is 1/(2pi) < 1/a and m < 1/a. Haven't figured out the inequalities between 1/(2pi) and m. Edit: Credits to an answer provided below, which I have incorporated.
Multiple choice
(https://i.imgur.com/0e0vjJE.png)
GeoGebra screenshot showing the tangent y=mx being barely above the curve y=cos(x) at x=a.
(https://i.imgur.com/csPKOtA.png)
Q17
(https://i.imgur.com/3vyuDGo.png)
Q25
According to the annuity table, an annuity of $1 at 0.75% interest, per annum compounding, for 8 years, will have a future value of $8.2132. Therefore an annuity of $1000 with everything else the same will have a future value of $8213.20.
Then we simply multiply (1.0125)2 for the future value after the extra 2 years, where no deposit is made, but the money still gains interest from just sitting there. Final answer is $8213.20 x 1.01252 = $8419.81 (nearest cent).
Q28-29
(https://i.imgur.com/BqL0ReW.png)
Q30-32
(https://i.imgur.com/F7xDGhk.png)
Q33-34
(https://i.imgur.com/o5Y9gqS.png)
I make no promises on every solution being accurate. Please point out mistakes and I'll get to them slowly.
Refer to fun_jirachi's solutions for questions I did not do.
11-14
(https://i.imgur.com/hdIPYYS.jpg)
15-16
(https://i.imgur.com/hftnuVv.jpg)
18-20
(https://i.imgur.com/kcc5H55.jpg)
21-24
(https://i.imgur.com/GlVnN1S.jpg)
26
(https://i.imgur.com/0a4RXIE.jpg)
27
(https://i.imgur.com/lGE9J5x.jpg)
Q10
Spoiler
For Q10 its 1/2pi < m < 1/a. The second inequality is because m=rise/run=e/a for some 0<e<1 (its not hard to see that 'a' is slightly to the left of 2pi, thus e=am is elightly below 1), so m = e/a < 1/a. The first inequality needs some explaining. Notice that the gradient of cos(x) decreases after x=a (just look in a small interval containing 2pi and a), so the tangent line is above the graph after x=a (at least until x=2pi). Consider the y value of the tangent line and cos(x) at x=2pi to get the inequality 1<m*2pi which implies 1/2pi < m.
(Another similar but graphical way) This value 1/2pi can be thought of as the gradient of a line L: y=x/2pi through (0,0) and (2pi,1). Remember that (2pi,1) lies on cos(x), so L will be below y=mx for x=2pi. So the gradient of L is smaller than that of y=xm, since they both pass through the origin. That is, 1/2pi < m.
The bit about the line being above cos(x) can be made more rigorous by considering their rate of change after x=a.