Answer:
Part A)
About $3767.34.
Part B)
About $3692.47.
Step-by-step explanation:
Part A)
Recall that compound interest is given by the formula:

Where <em>A</em> is the final amount, <em>P</em> is the initial amount, <em>r</em> is the interest rate, <em>n</em> is the number of times compounded per year, and <em>t</em> is the number of years.
To obtain $4000 after two years, let <em>A</em> = 4000 and<em> t</em> = 2.
Because the account pays 3% interest compounded monthly, <em>r</em> = 0.03 and <em>n</em> = 12.
Substitute and solve for <em>P: </em>
<em />
In concluion, about $3767.34 should be deposited.
Part B)
Recall the formula for continuous compound:

Where <em>e</em> is Euler's number.
Hence, let <em>A</em> = 4000, <em>r</em> = 0.04 and <em>t</em> = 2. Substitute and solve for <em>P: </em>
<em />
In conclusion, about $3692.47 should be deposited.