You are on the right track! First off, you should recognize that the sum of two independent poisson random variables (with parameters $ \lambda_1 $ and $ \lambda_2 $) is also poisson random variable (with parameter $ \lambda_1 + \lambda_2 $).
Let $ X_1 $ be the number of products delivered in 2 minutes from production line 1, and let $ X_2 $ be the number of products delivered in 2 minutes from production line 2. Since $ X_1 $ ~ $ Poisson(\lambda_1=1) $ and $ X_2 $ ~ $ Poisson(\lambda_2=1) $, then $ Y $ ~ $ Poisson(\lambda=\lambda_1+\lambda_2=2) $, where $ Y $ is the number of products delivered in 2 minutes from both production lines.
Finally, you can re-scale the poisson parameter to get "the number of products delivered in time t" as opposed to "the number of products delivered in 2 minutes." $ Z $ ~ $ Poisson(\lambda t) $, where $ t $ is given in increments of 2 minutes. It is often easier to use $ 1 $ minute as a unit of time; we can do this be letting $ t = \frac{1}{2} $, which gives us: $ Z $ ~ $ Poisson \big( \lambda t = (\lambda_1+\lambda_2)\frac{1}{2} = 1 \big) $, where $ Z $ is the number of products delivered in $ 1 $ minute from both production lines.
Part 1:
The "waiting time" of a Poisson distribution follows an exponential distribution with parameter $ \lambda $. The mean of an exponential distribution with parameter $ \lambda$ is equal to $ \frac{1}{\lambda} = \frac{1}{1} = 1 $.
Part 2:
$ P(Z > 2) = 1 - P(Z=0) - P(Z=1) - P(Z=2) = 1 - e^{-1} - e^{-1} - \frac{e^{-1}}{2} = 0.0803 $
Part 3:
The "waiting time of the 10th event" of a Poisson distribution is simply the sum of 10 independent exponential distributions with parameter $ \lambda $. Since the mean of an exponential distribution with parameter $ \lambda $ is $ \frac{1}{\lambda} = \frac{1}{1} = 1 $, then the mean of the sum of 10 independent exponential distributions (each with parameter $ \lambda $) is equal to $ 10 \times \frac{1}{\lambda} = 10 $.
Part 4:
Let $ t = 30 $ so that the unit of time in the Poisson distribution is $ 60 $ minutes = 1 $ hour $ (i.e. $ 30 \times 2 = 60 $ minutes). This means that we have: $ Z_1 $ ~ $ Poisson \big( \lambda t = (\lambda_1+\lambda_2)30 = 60 \big) $.
Since $ E[Z] = 60 $, Mary will receive (on average) $ 60 $ products in $ 1 $ hour. On average, Mary will not be able to wrap the last gift (as you pointed out in your question), so she will only wrap for $ 1,770 $ seconds (i.e. $ 59 $ gifts $ \times $ $ 30 $ seconds per gift $ = 1,770 $ seconds).
$ Percentage = \frac{1,770}{3,600} \times 100 \approx 48.36 \% $.
