Expand (4a2 + 12a + 9)3.5 up to fourth term, hence solve for (6.05)7

The expression $4a^2 + 12a + 9$ is a perfect square trinomial.

$$4a^2 + 12a + 9 = (2a)^2 + 2(2a)(3) + 3^2 = (2a + 3)^2$$

Now substitute this back into the original expression:

$$( (2a + 3)^2 )^{3.5} = (2a + 3)^{2 \times 3.5} = (2a + 3)^7$$

The binomial theorem states that $(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$.
Here, $x = 2a$, $y = 3$, and $n = 7$.

The first term ($k=0$):

$$T_1 = \binom{7}{0} (2a)^{7-0} (3)^0 = 1 \cdot (2a)^7 \cdot 1 = 128a^7$$

The second term ($k=1$):

$$T_2 = \binom{7}{1} (2a)^{7-1} (3)^1 = 7 \cdot (2a)^6 \cdot 3 = 7 \cdot 64a^6 \cdot 3 = 1344a^6$$

The third term ($k=2$):

$$T_3 = \binom{7}{2} (2a)^{7-2} (3)^2 = \frac{7 \times 6}{2 \times 1} \cdot (2a)^5 \cdot 9 = 21 \cdot 32a^5 \cdot 9 = 6048a^5$$

The fourth term ($k=3$):

$$T_4 = \binom{7}{3} (2a)^{7-3} (3)^3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \cdot (2a)^4 \cdot 27 = 35 \cdot 16a^4 \cdot 27 = 15120a^4$$

The expansion up to the fourth term is:

$$128a^7 + 1344a^6 + 6048a^5 + 15120a^4 + \dots$$

To solve for $(6.05)^7$, we can write it as $(6 + 0.05)^7$. This is in the form $(x+y)^n$ where $x=6$, $y=0.05$, and $n=7$. Since $y$ is a small value, the first few terms of the expansion will give a good approximation.

First term:

$$T_1 = \binom{7}{0} (6)^7 (0.05)^0 = 1 \cdot 279936 \cdot 1 = 279936$$

Second term:

$$T_2 = \binom{7}{1} (6)^6 (0.05)^1 = 7 \cdot 46656 \cdot 0.05 = 7 \cdot 2332.8 = 16329.6$$

Third term:

$$T_3 = \binom{7}{2} (6)^5 (0.05)^2 = 21 \cdot 7776 \cdot 0.0025 = 21 \cdot 19.44 = 408.24$$

Fourth term:

$$T_4 = \binom{7}{3} (6)^4 (0.05)^3 = 35 \cdot 1296 \cdot 0.000125 = 35 \cdot 0.162 = 5.67$$

$$(6.05)^7 \approx 279936 + 16329.6 + 408.24 + 5.67 = 296679.51$$

The expansion up to the fourth term is $128a^7 + 1344a^6 + 6048a^5 + 15120a^4$.
The value of $(6.05)^7$ is approximately 296679.51.

$$\boxed{\text{Expansion: } 128a^7 + 1344a^6 + 6048a^5 + 15120a^4; \text{ Value of } (6.05)^7 \approx 296679.51}$$