Discrete Time Fourier Transform to Frequency Domain
Most introductory texts on DSP have solved examples of this kind, so you should probably get some good text and work with it. But I'll give you a few hints anyway.

You know that

$$X(e^{j\Omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\Omega}\tag{1}$$

So for (a) you get

$$X(e^{j\Omega})=\sum_{n=-\infty}^{\infty}\left(\frac34\right)^n u[n-4] e^{-jn\Omega}\tag{2}$$

Since $u[n-4]$ equals zero for $n&lt;4$, (2) simplifies to

$$X(e^{j\Omega})=\sum_{n=4}^{\infty}\left(\frac34\right)^n e^{-jn\Omega}=
\sum_{n=0}^{\infty}\left(\frac34\right)^{n+4}e^{-j(n+4)\Omega}=\\=
\left(\frac34\right)^4e^{-j4\Omega}\sum_{n=0}^{\infty}\left(\frac34\right)^{n}e^{-jn\Omega}\tag{3}$$

where you can evaluate the final sum using the formula for a geometric series with $r=\frac34e^{-j\Omega}$.

For (b) you have

$$X(e^{j\Omega})=\sum_{n=-\infty}^{\infty}a^{|n|}e^{-jn\Omega}=\sum_{n=0}^{\infty}a^ne^{-jn\Omega}+\sum_{n=-\infty}^{-1}a^{-n}e^{-jn\Omega}=\\=
\sum_{n=0}^{\infty}a^ne^{-jn\Omega}+\sum_{n=1}^{\infty}a^{n}e^{jn\Omega}=
\sum_{n=0}^{\infty}a^ne^{-jn\Omega}+ae^{j\Omega}\sum_{n=0}^{\infty}a^{n}e^{jn\Omega}$$

Now you have both sums in the desired form to apply the formula:

$$X(e^{j\Omega})=\frac{1}{1-ae^{-j\Omega}}+\frac{ae^{j\Omega}}{1-ae^{j\Omega}}$$

You can combine the two terms, which should finally result in

$$X(e^{j\Omega})=\frac{1-a^2}{1-2a\cos\Omega+a^2}$$

Please remember that this is basic stuff that you must learn to do yourself. The only way to learn it is by solving the problems which you find at the end of each chapter of any introductory DSP text.