b) Without the yearly payment, the equation you used was derived using the present value of money:
$$P(1+r)^n = m \sum_{k=0}^{n-1} (1+r)^k$$
where $P$ is the principal and $r$ the monthly interest rate. If we allow a fixed yearly payment $f$ in addition, the equation becomes
$$P(1+r)^{12 y} = m' \sum_{k=0}^{12 y-1} (1+r)^k + f \sum_{k=0}^{y-1} (1+r')^k$$
where $y$ represents the number of years on the loan ($y=20$), and $r'$ is the effective annual rate of interest:
$$r' = (1+r)^{12}-1$$
Evaluating the geometric sums, we may solve for the new monthly payment $m'$:
$$m' = P \frac{r}{1-(1+r)^{-12 y}} - f \frac{r}{(1+r)^{12}-1}$$
Plugging in the numbers given above (i.e., $r=0.01$, $y=20$, $f=1000$,$P=50000$), we get a monthly payment of $m=471.69$.