略．

用这四项的后一项比前一项，如果能得到常数，则可证明它是等比数列．因为 $\dfrac{2^{a_{n+1}}}{2^{a_n}}=2^{a_{n+1}-a_n}=2^d$（$n=1,2,3$）是同一个常数，
所以 $2^{a_1}$，$2^{a_2}$，$2^{a_3}$，$2^{a_4}$ 依次构成等比数列．

不存在．理由略．

可设 $a_2=a$，将 $a_1,a_2,a_3,a_4$ 分别用 $a$ 和 $d$ 表示，然后假设 $a_1$，$a_2^2$，$a_3^3$，$a_4^4$ 依次构成等比数列的，于是得到关于 $a,d$ 的两个方程，化齐次解出 $\dfrac da$，从而推出矛盾．不存在．理由如下：
令 $a_1+d=a$，则 ${a_1}$，${a_2}$，${a_3}$，${a_4}$ 分别为 $a-d$，$a$，$a+d$，$a+2d$（$a>d$，$a>-2d$，$d\neq 0$）．
假设存在 $a_1$，$d$，使得 ${a_1}$，${a_2^2}$，${a_3^3}$，${a_4^4}$ 依次构成等比数列，则 $a^4=\left(a-d\right)\left(a+d\right)^3$，且 $\left(a+d\right)^6=a^2\left(a+2d\right)^4$．
令 $t=\dfrac da$，则 $1=\left(1-t\right)\left(1+t\right)^3$，且 $\left(1+t\right)^6=\left(1+2t\right)^4\left(-\dfrac 12<t<1,t\neq 0\right)$，化简得 $t^3+2t^2-2=0, \quad \cdots \cdots ① $
且 $t^2=t+1$．
将 $t^2=t+1$ 代入 $ ① $ 式，得 $t\left(t+1\right)+2\left(t+1\right)-2=t^2+3t=t+1+3t=4t+1=0$，则 $t=-\dfrac 14$．
显然 $t=-\dfrac 14$ 不是上面方程的解，矛盾，所以假设不成立，
因此不存在 $a_1$，$d$，使得 ${a_1}$，${a_2^2}$，${a_3^3}$，${a_4^4}$ 依次构成等比数列．

不存在，理由略．

首先将 $a_1,a_2,a_3,a_4$ 用 $a_1,d$ 表示，然后假设 $a_1^n$，$a_2^{n+k}$，$a_3^{n+2k}$，$a_4^{n+3k}$ 依次构成等比数列，于是得到关于 $a_1,d$ 的两个方程，齐次化得到关于 $\dfrac d{a_1}$ 的两个方程，可通过导数研究得出方程的解，通过解的不合理推出矛盾．不存在，理由如下：
假设存在 $a_1$，$d$ 及正整数 $n$，$k$，使得 $a_1^n$，$a_2^{n+k}$，$a_3^{n+2k}$，$a_4^{n+3k}$ 依次构成等比数列，则 $a_1^n\left(a_1+2d\right)^{n+2k}=\left(a_1+d\right)^{2\left(n+k\right)}$，且 $\left(a_1+d\right)^{n+k}\left(a_1+3d\right)^{n+3k}=\left(a_1+2d\right)^{2\left(n+2k\right)}$，分别在两个等式的两边同除以 $a_1^{2\left(n+k\right)}$ 及 $a_1^{2\left(n+2k\right)}$，并令 $t=\dfrac{d}{a_1}$（$t>-\dfrac 13$，$t\neq 0$），
则 $\left(1+2t\right)^{n+2k}=\left(1+t\right)^{2\left(n+k\right)}$，且 $\left(1+t\right)^{n+k}\left(1+3t\right)^{n+3k}=\left(1+2t\right)^{2\left(n+2k\right)}$．
将上述两个等式两边取对数，得 $\left(n+2k\right)\ln\left(1+2t\right)=2\left(n+k\right)\ln\left(1+t\right)$，且 $\left(n+k\right)\ln\left(1+t\right)+\left(n+3k\right)\ln\left(1+3t\right)=2\left(n+2k\right)\ln\left(1+2t\right)$．
化简得 $2k\left[\ln\left(1+2t\right)-\ln\left(1+t\right)\right]=n\left[2\ln\left(1+t\right)-\ln\left(1+2t\right)\right]$，且 $3k\left[\ln\left(1+3t\right)-\ln\left(1+t\right)\right]=n\left[3\ln\left(1+t\right)-\ln\left(1+3t\right)\right]$．
再将这两式相除，化简得 $\ln\left(1+3t\right)\ln\left(1+2t\right)+3\ln\left(1+2t\right)\ln\left(1+t\right)
=4\ln\left(1+3t\right)\ln\left(1+t\right). \quad \cdots \cdots ② $
令 $g\left(t\right)=4\ln\left(1+3t\right)\ln\left(1+t\right)-\ln\left(1+3t\right)\ln\left(1+2t\right)-3\ln\left(1+2t\right)\ln\left(1+t\right)$，则 $g'\left(t\right)=
\dfrac{2\left[\left(1+3t\right)^2\ln\left(1+3t\right)-3\left(1+2t\right)^2\ln\left(1+2t\right)+3\left(1+t\right)^2\ln\left(1+t\right)\right]}{\left(1+t\right)\left(1+2t\right)\left(1+3t\right)}$．
令 $\varphi\left(t\right)=\left(1+3t\right)^2\ln\left(1+3t\right)-3\left(1+2t\right)^2\ln\left(1+2t\right)+3\left(1+t\right)^2\ln\left(1+t\right)$，
则 $\varphi'\left(t\right)=6\left[\left(1+3t\right)\ln\left(1+3t\right)-2\left(1+2t\right)\ln\left(1+2t\right)+\left(1+t\right)\ln\left(1+t\right)\right]$，
令 $\varphi_1\left(t\right)=\varphi'\left(t\right)$，则 $\varphi_1'\left(t\right)=6\left[3\ln\left(1+3t\right)-4\ln\left(1+2t\right)+\ln\left(1+t\right)\right]$，
令 $\varphi_2\left(t\right)=\varphi_1'\left(t\right)$，则 $\varphi_2'\left(t\right)=\dfrac{12}{\left(1+t\right)\left(1+2t\right)\left(1+3t\right)}>0$．
由 $ g\left(0\right)=\varphi\left(0\right)=\varphi_1\left(0\right)=\varphi_2\left(0\right)=0 $，$ \varphi_2'\left(t\right)>0 $，
知 $ \varphi_2\left(t\right) $，$ \varphi_1\left(t\right) $，$ \varphi\left(t\right) $，$ g\left(t\right) $ 在 $ \left(-\dfrac 13,0\right) $ 和 $ \left(0,+\infty\right) $ 上均单调．
故 $ g\left(t\right) $ 只有唯一零点 $ t=0 $，即方程 $ ② $ 只有唯一解 $ t=0 $，故假设不成立．
所以不存在 $ a_1 $，$ d $ 及正整数 $ n $，$ k $，使得 $ a_1^n $，$ a_2^{n+k} $，$ a_3^{n+2k} $，$ a_4^{n+3k} $ 依次构成等比数列．