150. a) Find a four-digit number that is a perfect square and such that its first two digits are equal to each other and the last two digits are also equal to each other.

b) A two-digit number, when added to the number formed by the same digits in reverse order, results in a perfect square. Find all such numbers.
150. a) Let $a$ be the first and $b$ the last digit of the number $N$ we are looking for. Then this number is equal to $1000 a + 100 a + 10 b + b = 1100 a + 11 b = 11(100 a + b)$. Since the number $N$ is a perfect square, it follows from its divisibility by 11 that it is also divisible by 121, i.e., $\frac{N}{11} = 100 a + b$ is divisible by 11. But

$$
100 a + b = 99 a + (a + b) = 11 \cdot 9 a + (a + b)
$$

Therefore, $a + b$ is divisible by 11. Since both $a$ and $b$ are no greater than 9 and $a$ is not equal to 0, we have $1 \leqslant a + b \leqslant 18$, and thus $a + b = 11$.

From this,

$$
\begin{aligned}
100 a + b & = 11 \cdot 9 a + 11 = 11(9 a + 1) \\
\frac{N}{121} & = \frac{100 a + b}{11} = 9 a + 1
\end{aligned}
$$

Since $N$ is a perfect square, $\frac{N}{121}$ is also a perfect square. But among the numbers of the form $9 a + 1$, where $a$ ranges from 1 to 9, only $9 \cdot 7 + 1 = 64$ is a perfect square. Therefore, $N = 121 \cdot 64 = 7744 = 88^2$.

b) Let $a$ be the tens digit of this number, and $b$ the units digit. Then the number is $10 a + b$, and the number written with the same digits in reverse order is $10 b + a$. According to the problem's condition, $10 a + b + 10 b + a = 11(a + b) = k^2$.

From this, it follows that $k^2$ is divisible by 11, so $a + b$ is also divisible by 11. Since $a + b \leqslant 18$, this is only possible if $a + b = 11, k^2 = 121$. Thus, the numbers we are looking for are

$$
29, 38, 47, 56, 65, 74, 83, 92 \text{. }
$$