Apply recurrence formula:
$$ \begin{align}
a_{n+2} &= \sqrt{a_{n+1}\cdot a_n} \\
a_{n+2}^2 &= a_{n+1}\cdot a_n
\end{align}
$$
to compute explicitly:
$$ \begin{align}
a_3^2 &= a_2\cdot a_1, \quad \\
a_4^2 &= a_3\cdot a_2, \quad \\
a_5^2 &= a_4\cdot a_3, \quad \\
a_6^2 &= a_5\cdot a_4, \quad \\
\ldots &= \ldots \\
a_n^2 &= a_{n-1}\cdot a_{n-2}, \quad \\
a_{n+1}^2 &= a_n\cdot a_{n-1}, \quad \\
a_{n+2}^2 &= a_{n+1}\cdot a_n. \quad (1) \\
\end{align}$$
Multiply sides by sides:
$$ a_3^2\cdot a_4^2\cdot a_5^2 \cdot a_6^2\ldots a_n^2\cdot a_{n+1}^2\cdot a_{n+2}^2 = a_1\cdot a_2^2 \cdot a_3^2\cdot a_4^2\ldots a_{n-1}^2\cdot a_n^2\cdot a_{n+1}$$
Simplify:
$$ a_{n+1}\cdot a_{n+2}^2 = a_1\cdot a_2^2, \quad \forall n \ge 0. \quad (2) $$
For convenience, let's define:
$$c^2 = a_1a_2^2. \quad (3) $$
Then $(2)$ becomes:
$$ a_{n+2} = \frac{c}{\sqrt{a_{n+1}}} = \frac{c}{a_{n+1}^{1/2}}. \quad(4) $$
Apply $(4)$ to compute $a_{n+1}$ in terms of $a_n$:
$$ \begin{align}
a_{n+2} &= \frac{c}{ a_{n+1}^{1/2}}
= \frac{c}{ c^{1/2}{a_n^{-1/4}}} \\
&= c^{1/2} a_n^{1/2^2}. \quad (5)
\end{align}
$$
Thus, we can evaluate $(5)$ repeatedly for odd indices:
$$ \begin{align}
a_3 &= c^{1/2} \left( a_1 \right)^{1/2^2} \\
a_5 &= c^{1/2}\left( a_3 \right)^{1/2^2} = c^{1/2} \left( c^{1/2} a_1^{1/2^2} \right)^{1/2^n} = \left(c\right)^{2^{-1} + 2^{-3}} a_1^{2^{-4}} \\
a_7 &= c^{1/2}\left( a_5 \right)^{1/2^2} =
c^{1/2} \left( c^{1/2} a_3^{1/2^2} \right)^{1/2^n} = \left(c\right)^{2^{-1} + 2^{-3} + 2^{-5}} a_1^{2^{-6}} \\
\ldots &= \ldots \\
a_{2n+1} &= c^S a_1^{2^{-2n}}, \quad (6)
\end{align}$$
where $S$ is the following geometric series:
$$ S = \frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^5} \ldots + \frac{1}{2^{2n-1}}$$
$$\implies S - \frac{1}{2^2}S = \frac{1}{2} -\frac{1}{2^{2n+1}} $$
$$ \implies\lim_{n\to\infty} S = \frac{2}{3}.$$
Thus,
$$\lim_{n\to\infty} a_{2n+1} = c^{2/3} a_1^0 = \sqrt[3]{a_1a_2^2}. \quad (7)$$
Similarly, it can be shown that for even indices:
$$\lim_{n\to\infty} a_{2n+2} = c^{2/3} a_2^0 = \sqrt[3]{a_1a_2^2}. \quad (8)$$
Because both the odd and even sequences of $a_n$ are convergent to the same value, the sequence $a_n$ is convergent to the same value.
$$\lim_{n\to\infty} a_n = L = \sqrt[3]{a_1a_2^2}. \quad (9) $$
Thus, for the problem of interest, we have:
$$ L = \sqrt[3]{ab^2} = \sqrt[3]{1\cdot 8^2} = 4 $$
