Full Solution here for 2.33. The solution for 2.34 is really similar.

By Binomial Theorem, \[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + \binom{n}{n-1}x^{n-1} + \binom{n}{n}x^n. \] Let $\displaystyle\omega=\text{cis}\frac{2\pi}{3}$ be the imaginary cube root of unity, then $\omega^3=1$ and $1+\omega+\omega^2=0$, so \[ 1 + \omega^k + \omega^{2k} = \left\{ \begin{array}{lcl} 3, &\quad& \mbox{if } 3\mid k;\\ 0, &\quad& \mbox{if } 3\nmid k. \end{array} \right. \] Let $x=1, \omega, \omega^2$, and we get three equations, \[ \begin{array}{rcl} (1+1)^n &=& \displaystyle\binom{n}{0} + \binom{n}{1} + \binom{n}{2} +\cdots+ \binom{n}{n-1}+\binom{n}{n}, \\ (1+\omega)^n &=& \displaystyle\binom{n}{0} + \binom{n}{1}\omega + \binom{n}{2}\omega^2 +\cdots+ \binom{n}{n-1}\omega^{n-1}+\binom{n}{n}\omega^{n}, \\ (1+\omega^2)^n &=& \displaystyle\binom{n}{0} + \binom{n}{1}\omega^2 + \binom{n}{2}\omega^4 +\cdots+ \binom{n}{n-1}\omega^{2n-2}+\binom{n}{n}\omega^{2n}. \end{array} \] Multiplying the second and third equations by $\omega^2$ and $\omega^4$ respectively, \[ \begin{array}{rcl} (1+1)^n &=& \displaystyle\binom{n}{0} + \binom{n}{1} + \binom{n}{2} +\cdots+ \binom{n}{n-1}+\binom{n}{n}, \\ \omega^2(1+\omega)^n &=& \displaystyle\binom{n}{0}\omega^2 + \binom{n}{1}\omega^3 + \binom{n}{2}\omega^4 +\cdots+ \binom{n}{n-1}\omega^{n+1}+\binom{n}{n}\omega^{n+2}, \\ \omega^4(1+\omega^2)^n &=& \displaystyle\binom{n}{0}\omega^4 + \binom{n}{1}\omega^6 + \binom{n}{2}\omega^8 +\cdots+ \binom{n}{n-1}\omega^{2n+2}+\binom{n}{n}\omega^{2n+4}. \end{array} \] Adding these three equations, \[ 2^n + \omega^2(1+\omega)^n + \omega^4(1+\omega^2)^n = 3\left({\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\cdots}\right). \] We know that \[ \omega^2 = \text{cis}\frac{4\pi}{3}=\text{cis}\left({-\frac{2\pi}{3}}\right), \] \[ \omega^4 = \omega = \text{cis}\left({\frac{2\pi}{3}}\right), \] \[ 1+\omega=-\omega^2=-\text{cis}\frac{4\pi}{3}=\text{cis}\frac{\pi}{3}, \] and \[ 1+\omega^2=-\omega=\text{cis}\left({-\frac{\pi}{3}}\right), \] so \[ \begin{array}{cl} & 2^n + \omega^2(1+\omega)^n + \omega^4(1+\omega^2)^n \\ \\ =& \displaystyle 2^n + \text{cis}\left({-\frac{2\pi}{3}}\right)\text{cis}\frac{n\pi}{3} + \text{cis}\frac{2\pi}{3}\text{cis}\left({-\frac{n\pi}{3}}\right)\\ \\ =& \displaystyle 2^n + \text{cis}\frac{(n-2)\pi}{3} + \text{cis}\left({-\frac{(n-2)\pi}{3}}\right)\\ \\ =& \displaystyle 2^n + 2\cos\frac{(n-2)\pi}{3}, \end{array} \] therefore \[ \binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\cdots = \frac{1}{3}\left({2^n + 2\cos\frac{(n-2)\pi}{3}}\right). \]