A hexagon with all internal angles measuring $120^\circ$ can be inscribed in an equilateral triangle by extending three of its sides:

In both items, $x$, $y$, $z$ and $w$ are given. The equilateral triangle has sidelength $x + y + z$, so the remaining two sides of the hexagon have lengths $x + y + z - (w + z) = x + y - w$ and $x + y + z - (x + w) = y + z - w$. Thus the perimeter of the hexagon is $x + y + z + w + (x + y - w) + (y + z - w) = 2x + 3y + 2z - w$.

a.
We have $x = z = 5$, $y = 8$ and $w = 3$, so the answer is $2 \cdot 5 + 3 \cdot 8 + 2 \cdot 5 - 3 = 41$.

b.
Now $x = 3$, $y = 4$, $z = 5$ and $w = 1$, so the answer is $x + y - w + y + z - w = x + 2y + z - 2w = 3 + 2 \cdot 4 + 5 - 2 \cdot 1 = 14$.