Step 1: Condition for logarithm to exist.
For
\[
\log_2(x+3)
\]
to be defined,
\[
x+3\gt 0
\]
Thus,
\[
x\gt -3
\]
Step 2: Condition for square root in denominator.
Since the square root is in the denominator,
\[
x^2+3x+2\gt 0
\]
Factorizing:
\[
x^2+3x+2=(x+1)(x+2)
\]
Therefore,
\[
(x+1)(x+2)\gt 0
\]
This inequality is satisfied when:
\[
x\lt -2
\]
or
\[
x\gt -1
\]
Thus,
\[
x\in(-\infty,-2)\cup(-1,\infty)
\]
Step 3: Find the common interval.
From logarithm condition:
\[
x\gt -3
\]
From denominator condition:
\[
x\in(-\infty,-2)\cup(-1,\infty)
\]
Taking intersection:
\[
(-3,\infty)\cap\left[(-\infty,-2)\cup(-1,\infty)\right]
\]
we get:
\[
(-3,-2)\cup(-1,\infty)
\]
Step 4: Final conclusion.
Hence, the domain of the function is
\[
\boxed{(-3,-2)\cup(-1,\infty)}
\]