As I said in a comment, the amount of the loan should be calculated as $$\require{enclose}
L = 600 a^{(12)}_{\enclose{actuarial}{10} i} = 600\left(1-(1/1.01)^{120}\right)/0.01 = 41820.31
$$
You calculation of the total interest is correct. It's just the total payments less the amount of the loan.
The interest paid in the $10$ payment is $1\%$ of the amount of the loan outstanding after $9$ payments. Since there are then $111$ payments left, you do it it just as you calculated the amount of the loan, but with $111$ in the exponent instead of $120$.
For the amount of principal paid in the $20$ payment, calculate the amount of interest in the payment as above. The amount of principal in the payment is $600$ minus the amount of interest in the payment.
I haven't checked the arithmetic, but the logic looks right to me. I'd point out that the amount of interest in any payment is $jB$ so the amount of principal is $B(1−j)$, where $B$ is the loan balance just before the payment. This might be a little easier than computing two successive balances. Also, the answer is suspiciously close to $500,000$. You might check if that also gives $1860$.
EDIT
Let $j$ be the periodic interest rate, $j=.030301$. Let $v=\frac1{1+j}=.970590.$ Let $d$ be the discount rate, $d=\frac j{1+j}=0.02940985$ Now, just before the $29$th payment, there are $72$ payments remaining, and the first of them is due at once, so the loan balance is $$K\frac{1-v^{72}}{d}=30.03860564K$$ The interest in the payment is $j$ times the loan balance, or $.91019978$K, so that the principal in the payment is $.089800210K$
This gives $K=20712.646349$ and $$P=K\frac{1-v^{100}}i=649019.62$$
Best Answer
$$A = P \frac{r(1+r)^n}{(1+r)^n -1} \Rightarrow$$
$$A [(1+r)^n -1]= rP (1+r)^n \Rightarrow$$
$$(A-rP)(1+r)^n = A \Rightarrow$$
$$(1+r)^n = \frac{A}{A-rP}\Rightarrow$$
$$n = \log_{1+r} (\frac{A}{A-rP})$$