Answer
a) ${\bf 0.80} \; c$
b) ${\bf16}\;\rm year$
Work Step by Step
$$\color{blue}{\bf [a]}$$
Let's assume that $\rm S$ is Earth's frame and $\rm S′$ is the starship's frame.
The starship left the Earth for 26 years, one year of these years where when they were on the new planet.
So the time of the whole trip is 25 years.
Now we can find the speed of the starship since we know the distance between the two planets and the whole time of the journey back and forth.
$$v=\dfrac{2d}{t}$$
where $d$ is the distance between the two planets,
$$v=\dfrac{2(10c)}{25}= \color{red}{\bf 0.80} \; c$$
$$\color{blue}{\bf [b]}$$
The proper time measured by the starship is given by
$$t=\dfrac{\tau}{\sqrt{1-\frac{v^2}{c^2}}}$$
since we can deal with that two events happened at the same position (the Earth) within 25 years interval time.
$$\tau=t\sqrt{1-\frac{v^2}{c^2}}$$
Plug the known;
$$\tau=25\;\rm y\;\cdot\sqrt{1-\frac{0.8^2c^2}{c^2}}=\bf 15\;\rm year$$
Adding the year where they were settled on the planet, so the time measured by the ship is
$$ t'=\tau+1=15+1 =\color{red}{\bf16}\;\rm year$$
