a) $1038^2 = 1077444$.

b) By squaring a number which ends in 038 we get a number ending in 444, as shows the diagram:
$$\begin{array}{cccccc} \dots & \dots & \dots & 0 & 3 & 8 \\ \dots & \dots & \dots & 0 & 3 & 8 \\\hline \dots & \dots & \dots & 3 & 0 & 4 \\ \dots & \dots & \dots & 1 & 4 \\\hline \dots & \dots & \dots & 0 & & \\\hline \dots & \dots & \dots & 4 & 4 & 4 \end{array} \times$$
Since there are infinitely many numbers ending in 038, there are infinitely many perfect squares ending in 444.

c) Let $a$ – if possible – be a positive integer whose square ends in 4444. Then $a$ is even, that is $a = 2b$. Moreover, $a^2$ has form $10000k + 4444$, $k \in \mathbb{N}$, hence $4b^2 = a^2 = 4(2500k + 1111)$. It follows that the last two digits of $b^2$ are 1, 1, hence $b$ ends in 1 or 9.

The multiplications
$$\begin{array}{cccccc} \dots & \dots & \dots & n & 1 & \times \\ \dots & \dots & \dots & n & 1 & \\\hline \dots & \dots & \dots & n & 1 & \\\hline \dots & \dots & \dots & n & & \\\hline \dots & \dots & \dots & u & 1 & \end{array}$$
$$\begin{array}{cccccc} \dots & \dots & \dots & m & 9 & \times \\ \dots & \dots & \dots & m & 9 & \\\hline \dots & \dots & \dots & q & 1 & \\\hline \dots & \dots & \dots & p & & \\\hline \dots & \dots & \dots & v & 1 & \end{array}$$
show that digits $u$ and $v$ are even: $u$ is the last digit of $2n$, $q$ is the last digit of $p + 8$ and $v$ is the last digit of the sum $2p + 8$.
Therefore, the next to the last digit of $b^2$ is even, so it can not be 1. This shows that $a$ can not exist.