13. Given the function $f(x)=\frac{1}{\lg \left(2^{x}+4 \cdot 2^{-x}-a\right)}$ with the domain $\mathbf{R}$, find the range of the real number $a$.

 untranslated part:
 已知函数 $f(x)=\frac{1}{\lg \left(2^{x}+4 \cdot 2^{-x}-a\right)}$ 的定义域为 $\mathbf{R}$, 求实数 $a$ 的取值范围。

translated part:
Given the function $f(x)=\frac{1}{\lg \left(2^{x}+4 \cdot 2^{-x}-a\right)}$ with the domain $\mathbf{R}$, find the range of the real number $a$.
13. From the problem, we have $\lg \left(2^{x}+4 \cdot 2^{-x}-a\right) \neq 0$ for all $x \in \mathbf{R}$, which means 
$$
\left\{\begin{array}{l}
2^{x}+4 \cdot 2^{-x}-a>0 \\ 
2^{x}+4 \cdot 2^{-x}-a \neq 1 \text {. }
\end{array}\right.
$$ 
Therefore,
$$
\left\{\begin{array}{l}
a<2^{x}+4 \cdot 2^{-x} \\
a \neq 2^{x}+4 \cdot 2^{-x}-1
\end{array}\right.
$$

Since $2^{x}+4 \cdot 2^{-x} \geqslant 4$, from (1) we know $a<4$, and from (2) $a \neq 2^{x}+4 \cdot 2^{-x}-1 \geqslant 3$, thus $a<3$.