Hint: The outstanding balance after $n$ quarterly made payments is
$$B_n=1000\cdot \left( 1+\frac{0.1}4 \right)^n-100\cdot \frac{\left( 1+\frac{0.1}4 \right)^n-1}{\frac{0.1}4 }$$
Therefore the outstanding balance is less than 500, if $B_n<500$. You have to solve it for $n$. If we multiply the inequality by $\frac{0.1}4 =\frac1{40}$ and substitute $1+\frac{0.1}4$ by $q$ we obtain
$$25\cdot q^n-100\cdot \left( q^n-1\right)<12.5$$
Remark:
In this context I can only refer to problems where a loan ($C_0$) is taken out and the repayments are constant. If the repayments are made $m$ times in a year and the (constant) repayments (r) are made at the end of each period we have the following term:
$$B_n=C_0\cdot \left( 1+\frac{i}m \right)^{n}-r\cdot \frac{\left( 1+\frac{i}m \right)^{n}-1}{\frac{i}m }$$
$B_n$: Outstanding balance after $n$ periods.
$i$: nominal interest rate
Usually we have a constant interest rate i and a constant repayment (annuity). In this case we can set $B_n$ equal to zero and we are able to calculate the missing value of $i, n,r$ or $C_0$
If the repayments are made at the $\color{blue}{\textrm{beginning}}$ of each period the formula changes to
$$B_n=C_0\cdot \left( 1+\frac{i}m \right)^{n}-r\cdot \color{blue}{ \left( 1+\frac{i}m \right)}\cdot \frac{\left( 1+\frac{i}m \right)^{n}-1}{\frac{i}m }$$
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.
Best Answer
Conceptually, the way amortization of loan payments works is that the lender should come out even at the end of the loan when you account for future value of money based on the loan's interest rate. If the lender loaned $A$ dollars at an interest rate per period $r$ over $n$ periods, the future value (at the end of the loan) of the money lent is $A(1+r)^n$. If the payment is $p$, then the future value of the first payment is $p(1+r)^{n-1}$ (paid at the end of the first period), for the second is $p(1+r)^{n-2}$, for the $k$th payment is $p(1+r)^{n-k}$, up to the last payment which is paid at the end of the loan so its future value is $p(1+r)^0=p$. If you take the sum of these future values of all the payments, it should be equal to the future value of the original loan. This equation can be solved for $p$ in terms of the general $A$, $r$, and $n$ to get a generalized amortization formula.
Now, on a period-to-period basis, the balance is defined by the initial balance $b_0=A$ and the recurrence relation $b_{k}=(1+r)b_{k-1}-p$. That is, the change from one period to the next is to add the interest due and subtract the payment. Knowing the parameters of the loan and the payment, you can use this to find the balance after a specific number of periods, as well as the breakdown of interest and principal in each payment.