I'm not sure if this is what you're looking for, but you can use the delta method to approximate the standard error of $\beta_{1}+\beta_{2}$:
$$\newcommand{\SE}{\operatorname{SE}}\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}
\SE(\beta_{1}+\beta_{2})\approx \sqrt{\Var(\beta_{1})+\Var(\beta_{2})+2\cdot \Cov(\beta_{1},\beta_{2})}
$$
After the regression command in R, you can type vcov(model)
which gives you the variance-covariance matrix of the coefficients. The values on the diagonal of the variance-covariance matrix are the variances of the respective coefficients while the values off-diagonal represent the covariances between the corresponding coefficients.
With that you can calculate the confidence interval and the $t$-value (for a Wald-test):
$$
t_{\beta_{1}+\beta_{2}}=\frac{(\beta_{1}+\beta_{2} ) - 1}{\SE(\beta_{1}+\beta_{2})}
$$
And from that you can use the $t$-distribution to calculate a two- or one-sided $p$-value with 2*pt(-abs(t), df=n-1)
for a two-sided $p$-value and pt(-abs(t), df=n-1)
($\leq 1$) and 1-pt(-abs(t), df=n-1)
($\geq 1$) for the one-sided $p$-values. Please note that the $t$-distribution and the $F$-distribution are closely related: the square of the $t$-distribution with df degrees of freedom is the $F$-distribution with 1 numerator degree of freedom and df denominator degrees of freedom. It doesn't matter if you use the $t$-value and the $t$-distribution or the squared $t$-value and the $F$-distribution (pf(t^2, df1=1, df2=n-1)
) to calculate the $p$-values. So I don't know what the difference between 1) and 2) is.
EDIT:
The R
-code for the steps explained above are (it's not the fastest way but I concentrated on legibility):
1) Calculate the variance-covariance-matrix with
vcov.mat <- vocv(cobbdoug)
2) Calculate the approximate standard error of $\beta_{1}+\beta_{2}$:
se.b1b2 <- sqrt(vcov.mat[2,2] + vcov.mat[3,3] + 2*vcov.mat[2,3])
3) Calculate the $t$-value:
t.val <- ((coef(cobbdoug)[2] + coef(cobbdoug)[3]) - 1)/se.b1b2
4) Calculate the $p$-value for question 1):
2*pt(-abs(t.val), df=cobbdoug$df.residual) # assuming that you have 15 years and 3 coefficients, so df=15-3 = 12
5) Calculate the $p$-value for question 3):
pt(-abs(t.val), df=cobbdoug$df.residual)