Taking the base-2 $\log$ of the sequence $\left\{a_{n}\right\}$ converts the multiplicative rule to a more familiar additive rule: $\log _{2}\left(a_{n+1}\right)-4 \log _{2}\left(a_{n}\right)+5 \log _{2}\left(a_{n-1}\right)-2 \log _{2}\left(a_{n-2}\right)=0$. The characteristic equation is $0=x^{3}-4 x^{2}+5 x-2=(x-1)^{2}(x-2)$, so $\log _{2}\left(a_{n}\right)$ is of the form $a \cdot n+b+c \cdot 2^{n}$ and we find $a_{n}=2^{2 n+2^{n-1}}$. Now, $$\sqrt{a_{1}+\sqrt{a_{2}+\sqrt{a_{3}+\cdots}}}=\sqrt{2} \cdot \sqrt{4+\sqrt{16+\sqrt{64+\cdots}}}$$ We can estimate the new nested radical expression as 3, which expands thus $$3=\sqrt{4+5}=\sqrt{4+\sqrt{16+9}}=\sqrt{4+\sqrt{16+\sqrt{64+17}}}=\cdots$$ As a rigorous confirmation, we have $2^{k}+1=\sqrt{4^{k}+\left(2^{k+1}+1\right)}$, as desired. It follows that the answer is $3 \sqrt{2}$.