Apply the binomial theorem.

The binomial theorem states that $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.
For $(x+2y)^5$, we have $a=x$, $b=2y$, and $n=5$.
The binomial coefficients for $n=5$ are $\binom{5}{0}=1$, $\binom{5}{1}=5$, $\binom{5}{2}=10$, $\binom{5}{3}=10$, $\binom{5}{4}=5$, $\binom{5}{5}=1$.

The binomial expansion is $\boxed{x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5}$.

We can write 1.02 as $1 + 0.02$.
Comparing $(x+2y)^5$ with $(1+0.02)^5$, we can set $x=1$ and $2y=0.02$.
From $2y=0.02$, we get $y=0.01$.

Calculate each term:

Rounding to three decimal places, we look at the fourth decimal place. Since it is 0 (which is less than 5), we keep the third decimal place as it is.

The value of $(1.02)^5$ correct to three decimal places is $\boxed{1.104}$.