Part d)
There are $4^n$ possible values for $(x_1, x_2, \ldots, x_{2n-1}, x_{2n})$ which we can divide into $n$ $2$-bit vectors $(x_1,x_2), (x_3,x_4),\ldots, (x_{2n-1},x_{2n})$ each of which can take on $4$ values. Then,
$x_{1}x_{2}\vee x_{3}x_{4}\vee\ ...\ \vee\ x_{2n-1}x_{2n}= 0$ if and only if $(x_1,x_2) \neq (1,1)$ and $(x_3,x_4) \neq (1,1)$ and $\cdots$ and $(x_{2n-1},x_{2n}) \neq (1,1)$. Thus, $3^n$ of the $4^n$ values of $(x_1, x_2, \ldots, x_{2n-1}, x_{2n})$ result in the expression above having value $0$.
Thus, the number of solutions to $x_{1}x_{2}\vee x_{3}x_{4}\vee\ ...\ \vee\ x_{2n-1}x_{2n}= 1$ is $4^n-3^n$.
Part c)
Let $x_{2i-1}x_{2i} = y_i$ where we think of $Y_i$ as a Bernoulli random variable with parameter $p = \frac{1}{4}$, and $Z = Y_1+Y_2+\cdots+Y_{n}$ as a binomial random variable with parameters $(n,p)$. Now, the value of $Z$ is an odd number if and only if $x_{1}x_{2}\oplus x_{3}x_{4}\oplus\ ...\ \oplus\ x_{2n-1}x_{2n}=1$ and we have that $\begin{align*}P\{Z ~\text{is an odd number}\} &= \binom{n}{1}p(1-p)^{n-1}+\binom{n}{3}p^3(1-p)^{n-3} + \cdots\\ &= \frac{1}{2}\left[\sum_{j=0}^n \binom{n}{j}p^j(1-p)^{n-j} - \sum_{j=0}^n \binom{n}{j}(-1)^jp^j(1-p)^{n-j}\right]\\ &= \frac{1}{2}\left[(p + (1-p))^n - ((1-p)-p)^n\right]\\ &= \frac{1}{2}\left[1 - (1-2p)^n\right]\\ &= \frac{1}{2} - \frac{1}{2^n}.\end{align*}$ Turning back to Boolean variables, we conclude that
The number of solutions to $x_{1}x_{2}\oplus x_{3}x_{4}\oplus\ ...\ \oplus\ x_{2n-1}x_{2n}=1$ is $2^{2n-1}-2^{n-1}$.
I will leave it to you to try and apply these ideas to parts a) and b)