Answer
Both of the investments are almost the same, but to be more precise, (a) is a better investment.
Work Step by Step
*A quick review*
$A = P(1+\frac{r}{n})^{nt}$
$n$ - number of times investment is compounded per year
$t$ - number of years
$r$ - interest rate per year (In decimal form)
$P$ - principal
$A$ - amount of money after $t$ years
For continuously compounded interest:
$A=Pe^{rt}$
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To determine which of the investments is the best we can take any arbitrary number for principal and number of years. Let's say $P=\$1000$ and $t=1$
(a) $A=1000(1+\frac{0.05125}{2})^{2\times1}=1000\times1.025625^2\approx\$1051.90$
(b) $A=1000e^{0.05\times1}\approx\$1051.27$
Although it is nearly the same, (a) is still a little bit better investment
