Dividing through by $L$ gives you output per worker:
$$\frac{Y}{L}= \frac{A L^{\alpha}K^{\beta }}{L}= \frac{A L^{\alpha}K^{\beta }}{L^\alpha L^\beta}=A \left( \frac{K}{L}\right)^\beta.$$
We used the first assumption. Now take logs, which gets you
$$\ln \frac{Y}{L}= \ln A + \beta \cdot \ln \frac{K}{L}.$$
Calculate the difference from adjusting $K$ and $L$:
$$\ln \frac{Y}{L}-\ln \frac{Y'}{L'}= \ln A-\ln A' + \beta \left( \ln \frac{K}{L}-\ln \frac{K'}{L'} \right).$$
For small changes, log differences are approximately equal to growth rates*, which leaves you with one equation in one unknown:
$$ 3\%= \% \Delta A + 0.4 \cdot 4\%.$$
This uses assumptions 2-4.
*Take a Taylor series expansion of $\ln(1+x)$ to convince yourself that this is true if this is not something that you're intimately familiar with by now.
One method you can apply is the lagrange-multiplier method. The lagrange function is:
$\mathcal L=Q(L,K)+\lambda (B-C(L,K))$
$\mathcal L=L^{0.5}\cdot K^{0.5}+\lambda (800-25L-40K)$
The partial dervatives, w.r.t L and K, are:
$\frac{\partial \mathcal L}{\partial L}=0.5\cdot L^{-0.5}\cdot K^{0.5}-\lambda\cdot 25=0 $
$\frac{\partial \mathcal L}{\partial K}=0.5\cdot L^{0.5}\cdot K^{-0.5}-\lambda\cdot 40=0 $
Bringing the terms with $\lambda$ on the RHS
$0.5\cdot L^{-0.5}\cdot K^{0.5}=\lambda\cdot 25 $
$0.5\cdot L^{0.5}\cdot K^{-0.5}=\lambda\cdot 40 $
Dividing the first equation by the second equation:
$\Large{\frac{0.5\cdot L^{-0.5}\cdot K^{0.5}}{0.5\cdot L^{0.5}\cdot K^{-0.5}}=\frac{\lambda\cdot 25}{\lambda\cdot 40}}$
Cancelling out 0.5 and $\lambda$
$\Large{\frac{ L^{-0.5}\cdot K^{0.5}}{ L^{0.5}\cdot K^{-0.5}}=\frac{ 25}{ 40}}$
$\frac{ L^{-0.5}}{ L^{0.5}}=L^{-0.5}\cdot L^{-0.5}=L^{-0.5-0.5}=L^{-1}=\frac{1}{L}$
$\frac{ K^{0.5}}{ K^{-0.5}}=K^{0.5}\cdot K^{-(-0.5)}=K^{0.5+0.5}=K^{1}=K$
This gives for the RHS $\frac{K}{L}$
$\frac{K}{L}=\frac{ 25}{ 40} \ \ \Rightarrow \ \ K=\frac{5}{8}L \quad \color{blue}{(*)}$
Partial derivative w.r.t $\lambda$
$\frac{\partial \mathcal L}{\partial \lambda}=800-25L-40K=0$
Now you can insert the expression for K, which is in $\color{blue}{(*)}$.
$800-25L-40\frac{5}{8}L=0$
$800-25L-25L=0 \ \ \Rightarrow \ \ 800=50L \ \ \Rightarrow \ \ L^*=16$
Finally you can use $\color{blue}{(*)}$ again to calculate $K^*$
Best Answer
It is a normalisation factor intended to make $U(1-\alpha,\alpha)=1$ and is not necessary for the definiton of Cobb-Douglas form. In fact, the most general way to define Cobb-Douglas production function is as follows:
$$U(x_1,x_2,\ldots, x_n)=\Pi_{i} {x_i}^{\lambda_i} \text{ where }\sum_i \lambda_i=1$$