Specific steps in applying the Chinese Remainder Theorem to solve modular problem splitting modulus

Welcome to Math SX! You have to use Euler's theorem as $\varphi(4)=2$, $\;\varphi(25)=20$ we have $$ 7^{30}\equiv7^{30\bmod2}=1\mod 4,\qquad 7^{30}\equiv7^{30\bmod20}=7^{10}\mod 25$$ To find the latter power, you can use the modular fast exponentiation algorithm, but here, it will be simpler: modulo $25$, $$7^2\equiv -1\enspace\text{so}\enspace 7^4=1,\enspace\text{hence } \;7^{30}\equiv 7^{30\bmod 4}=7^2\equiv -1.$$ Finally, since $\;25-6\cdot 4=1$ (Bézout's identity), $$7\equiv \begin{cases}\phantom{-}1\mod4\\-1\mod 25\end{cases}\iff 7\equiv 1\cdot 25-(-1)\cdot 6\cdot 4=49\mod 100.$$

$$7^{30}\equiv x\pmod{ 100}$$

Could be solved easily without Chinese Remainder Theorem. Note that $7^4=2401 \equiv 1\pmod {100} $ Thus $$ 7^{30} = 7^{28}\times 49 \equiv 49 \pmod {100}$$

Solving the system with Chinese Remainder Theorem requires finding a linear combination of $25$ and $4$ to equal 1.

Such a combination is $$ 1= 1(25) -6(4) $$

Therefore the answer to the system is $$ x\equiv (1)(1)(25) +(-1)(-6)(4) \pmod {100}$$

That is $$ x\equiv 49 \pmod {100}$$

Chinese Remainder Theorem says:

$$\mathbb{Z}/100 \simeq \mathbb{Z}/25 \times \mathbb{Z}/4$$

where the isomorphism is given by mapping $x \pmod {100}$ to $(x \pmod {25}, x \pmod {4})$. Thus, the class of $x$ is a number from $0$ to $99$ that congruence to $1 \bmod 4$ and $24 \bmod 25$. The numbers $0 \le x \le 99$ and $x \equiv 24 \pmod {25}$ are: $24, 49, 74, 99$. Now which one is congruence to $1 \bmod 4$?