This is the reverse of your question; you asked what happens to the value of a continued fraction whose terms are modified in a simple way, but you also implied that you were interested in what to do to the terms to effect simple operations on the value of the fraction.
There is an algorithm which, given a single continued fraction $x$, will give you back the continued fraction $ax+b\over cd+d$ for any fixed integers $a,b,c,d$.
There is an analogous algorithm which takes any two continued fractions $x$ and $y$ and yields the continued fraction for $axy+bx+cy+d\over exy+fx+gy+h$ for any fixed integers $a,\ldots, h$. By putting $\langle a,\ldots,h\rangle = \langle 1,0,0,0,0,0,0,1\rangle$, the algorithm calculates the continued fraction for $xy$; by putting $\langle a,\ldots,h\rangle = \langle 0,1,1,0,0,0,0,1\rangle$, the algorithm calculates the continued fraction for $x+y$.
These algorithms were first discovered in the 1970s by Bill Gosper. They are not too complicated, but they are not trivial either. To explain them in this forum would probably not be very productive.
Gosper's monograph is available here.
I have some slides for a talk on the subject that I gave at Haverford College a few years ago, and an implementation in C. I would not normally have chosen C for this work, but I wanted to make the point that the algorithm does not require the fancy features of language $X$ for any $X$.
I also recently corresponded with Art DuPre of Rutgers University, who, with Dave Reimer of the College of New Jersey recently discovered algorithms which take a continued fraction $x$ and calculates the continued fraction for $\frac12x$. I don't know that these are published yet, and I haven't looked at them closely.
Addendum: I took another look at the Dupre-Reimer algorithm. It is a mess. It is also unaccompanied by a proof, and I imagine that the easiest way to prove it would be to show that it synthesizes the same results as the Gosper algorithm, which seems simpler.
Best Answer
Not quite. The recursive formulas require the previous two terms, and so you need two base cases to start them off. (In other words: how did you decide that $A_1=3$ and $B_1=3$ given the data in your post?)
Here, in addition to $A_0=q_0$ and $B_0=1$, you can use $A_{-1} = 1$ and $B_{-1}=0$. This leads in part (a) to $A_0 = 1, A_1 = 3, A_2 = 10, A_3 = 43$ as you obtained, but instead to $B_0 = 1, B_1 = 2, B_2 = 7, B_3=30$, so that the fraction is $\frac{43}{30}$.
You can check that this is the correct answer by carrying out the Euclidean algorithm on $43$ and $30$ and recording the partial quotients; they are indeed 1, 2, 3, 4.